Problem

Quantum Mechanics Problems David J. Jeffery Department of Physics & Health Physics Idaho State University Pocatello, Idaho

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Portpentagram Publishing (self-published) 2001 January 1

Introduction Quantum Mechanics Problems (QMP) is a source book for instructors of introductory quantum mechanics. The book is available in electronic form to instructors by request to the author. It is free courseware and can be freely used and distributed, but not used for commercial purposes. The aim of QMP is to provide digestable problems for quizzes, assignments, and tests for modern students. There is a bit of spoon-finding—nourishing spoon-feeding I hope. The problems are grouped by topics in chapters: see Contents below. The chapter ordering follows roughly the traditional chapter/topic ordering in quantum mechanics books. For each chapter there are two classes of problems: in order of appearance in a chapter they are: (1) multiple-choice problems and (2) full-answer problems. Almost all the problems have complete suggested answers. The answers may be the greatest benefit of QMP. The questions and answers can be posted on the web in pdf format. The problems have been suggested by many sources, but have all been written by me. Given that the ideas for problems are the common coin of the realm, I prefer to call my versions of the problems redactions. Instructors, however, might well wish to find solutions to particular problems from well known texts. Therefore, I give the suggesting source (when there is one or when I recall what it was) by a reference code on the extra keyword line: e.g., (Gr-10:1.1) stands for Griffiths, p. 10, problem 1.1. Caveat: my redaction and the suggesting source problem will not in general correspond perfectly or even closely in some cases. The references for the source texts and other references follow the contents. A general citation is usually, e.g., Ar-400 for Arfken, p. 400. At the end of the book are three appendices. The first is set of review problems anent matrices and determinants. The second is an equation sheet suitable to give to students as a test aid and a review sheet. The third is a set of answer tables for multiple choice questions. Quantum Mechanics Problems is a book in progress. There are gaps in the coverage and the ordering of the problems by chapters is not yet final. User instructors can, of course, add and modify as they list. Everything is written in plain TEX in my own idiosyncratic style. The questions are all have codes and keywords for easy selection electronically or by hand. A fortran program for selecting the problems and outputting them in quiz, assignment, and test formats is also available. Note the quiz, etc. creation procedure is a bit clonky, but it works. User instructors could easily construct their own programs for problem selection. I would like to thank the Department of Physics & Health Physics of Idaho State University for its support for this work. Thanks also to the students who helped flight-test the problems.

Contents 1 2 3 4 5 6 7 8 9 10 11

Classical Physics in Trouble QM Postulates, Schr¨odinger Equation, and the Wave Function Infinite Square Wells and Other Simple Wells The Simple Harmonic Oscillator (SHO) Free Particles and Momemtum Representation Foray into Advanced Classical Mechanics Linear Algebra Operators, Hermitian Operators, Braket Formalism Time Evolution and the Heisenberg Representation Measurement The Central Force Problem and Orbital Angular Momentum i

12 13 14 15 16 17 18 19 20 21 22 23 24 25

The Hydrogenic Atom General Theory of Angular Momentum Spin Time-Independent Approximation Methods Time-Dependent Perturbation The Hydrogenic Atom and Spin Symmetrization Principle Atoms Molecules Solids Interaction of Radiation and Matter Second Quantization Klein-Gordon Equation Ephemeral Problems

Appendices 1 Mathematical Problems 2 Quantum Mechanics Equation Sheet 3 Multiple-Choice Problem Answer Tables

References Abramowitz, M., & Stegun, I. A. 1972, Handbook of Mathematical Functions with Formulas, Graphs,s and Mathematical Tables (Washington, D.C.: U.S. Government Printing Office) (AS) Adler, R., Bazin, M., & Schiffer, M. 1975, Introduction to General Relativity (New York: McGraw-Hill Book Company) (ABS) Arfken, G. 1970, Mathematical Methods for Physicists (New York: Academic Press) (Ar) Baym, G. 1969, Lectures on Quantum Mechanics (Reading, Massachusetts: Benjamin/Cummings Publishing Co., Inc.) (Ba) Behte, H., & Jackiw, R. W. 1986, Intermediate Quantum Mechanics (Menlo Park, California: Benjamin/Cummings Publishing Company) (BJ) Bernstein, J., Fishbane, P. M., & Gasiorowicz, S. 2000, Modern Physics (Upper Saddle River, New Jersey: Prentice Hall) (BFG) Bevington, P. R. 1969, Data Reduction and Error Analysis for the Physical Sciences (New-York: McGraw-Hill Book Company) (Bev) Cohen-Tannoudji, C., Diu, B., & Lalo¨e, F. 1977, Quantum Mechanics (New York: John Wiley & Sons) (Co) Chandrasekhar, S. 1960, Radiative Transfer (New York: Dover Publications, Inc.) (Ch) Dahl, J. P. 2001, Introduction to the Quantum World of Atoms and Molecules (Singapore: World Scientific) (Da) Davisson, C. M. 1965, in Alpha-, Beta-, and Gamma-Ray Spectroscopy, ed. K. Siegbahn (Amsterdam: North-Holland), 37 (Da) Enge, H. A. 1966, Introduction to Nuclear Physics (Reading, Massachusetts: Addison-Wesley Publishing Company, Inc.) (En) Gasiorowicz, S. 1974, Quantum Physics (New York: John Wiley & Sons) (Ga) Greiner, W. 1994, Quantum Mechanics: An Introduction (Berlin: Springer-Verlag) (Gre) Griffiths, D. J. 1995, Introduction to Quantum Mechanics (Upper Saddle River, New Jersey: Prentice Hall) (Gr) Griffiths, D. J. 2005, Introduction to Quantum Mechanics (Upper Saddle River, New Jersey: Prentice Hall) (Gr2005) ii

Goldstein, H., Poole, C. P., Jr., & Safko, J. L. 2002, Classical Mechanics, 3rd Edition (San Francisco: Addison-Wesley Publishing Company) (GPS) Harrison, W. A. 2000, Applied Quantum Mechanics (Singapore: World Scientific) (Ha) Hodgman, C. D. 1959, CRC Standard Mathematical Tables, 12th Edition (Cleveland, Ohio: Chemical Rubber Publishing Company) (Hod) Jackson, J. D. 1975, Classical Electrodynamics (New York: John Wiley & Sons) Jeffery, D. J. 2001, Mathematical Tables (Port Colborne, Canada: Portpentragam Publishing) (MAT) Keenan, C. W., Wood, J. H., & Kleinfelter, D. C., 1976, General College Chemistry (New York: Harper & Row, Publishers) (Ke) Leighton, R. B. 1959, Principles of Modern Physics (New York: McGraw-Hill Book Company, Inc.) (Le) Mihalas, D. 1978, Stellar Atmospheres, 2nd Edition (San Francisco: W. H. Freeman and Company) (Mi) Morrison, M. A. 1990, Understanding Quantum Mechanics (Upper Saddle River, New Jersey: Prentice Hall) (Mo) Morrison, M. A., Estle, T. L., & Lane, N. F. 1991, Understanding More Quantum Mechanics (Upper Saddle River, New Jersey: Prentice Hall) (MEL) Pathria, R. K. 1980, Statistical Mechanics (Oxford: Pergamon Press) (Pa) Pointon, A. J. 1967, Introduction to Statistical Physics (London: Longman Group Ltd.) (Po) Tipler, P. A. 1978, Modern Physics (New York: Worth Publishers, Inc.) (Ti)

iii

Chapt. 1 Classical Physics in Trouble

Multiple-Choice Problems 001 qmult 00100 1 1 3 easy memory: quantum mechanics 1. The physical theory that deals mainly with microscopic phenomena is: a) b) c) d) e)

quartz mechanics. quarks mechanics. quantum mechanics. quantum jump mechanics. quasi-mechanics.

001 qmult 00200 1 1 1 easy memory: photon energy 2. The photon, the quantum of electromagnetic radiation, has ENERGY: a) b) c) d) e)

hf = − h ω. h/λ. − h k. h2 f . hf 2 .

001 qmult 00300 1 1 4 easy memory: photoelectric effect 3. A key piece of evidence for the wave-particle duality of light is: a) b) c) d) e)

the the the the the

photograph effect. Maxwell’s electrodynamics as summarized in the four Maxwell’s equations. frequency of red light. photoelectric effect. photomagnetic effect.

001 qmult 00400 1 1 1 easy memory: Compton effect 4. Einstein predicted and Compton proved that photons: a) b) c) d) e)

have linear momentum. do not have linear momentum. sometimes have linear momentum. both have and do not have linear momentum at the same time. neither have nor have not linear momentum.

001 qmult 00500 1 4 3 easy deducto-memory: Bohr atom 5. “Let’s play Jeopardy! For $100, the answer is: This model of an atom is of historical and pedagogical interest, but it is of little use in modern practical calculations and from the modern standpoint is probably misleading rather than insight-giving.” What is

, Alex? 1

2

Chapt. 1 Classical Physics in Trouble a) Schr¨odinger’s model of the hydrogen atom b) the Thomas-Fermi model of a many electron atom c) Bohr’s model of the hydrogen atom d) the liquid drop model of the atom e) the model hydrogen atom of Leucippos and Democritos

001 qmult 00550 1 1 4 easy memory: hydrogenic energy formula 6. The formula Z2 1 En = − me c2 α2 2 2 n gives the main energy levels of: a) b) c) d) e)

positronium. magnesium deboride. the hydrogen molecule. the hydrogenic atom. the infinite square well.

001 qmult 00600 1 1 5 easy memory: Greek atomists 7. The atomic theory was first proposed by the ancient Greeks Leucippos (5th century BCE) and Democritos (5th to 4th century BCE: he reputedly lived to be 100). The term atomos means uncut: e.g., the grass is atomos. The atomists started from a philosophical position that there had to be something to give stability to nature: obviously the macroscopic world was full of change: therefore what was imperishable or uncutable—atoms—must be below perception. The modern quantum theory does indeed bear out some of their thinking. Microscopic particles can be created and destroyed, of course, but the members of a class are much more identical than macroscopic objects can ever be: fundamental particles like electrons and quarks are thought to be absolutely identical. Thus the forms particles can take are apparently eternal: a hydrogen atom today is the same in theory as one at any time in universal history. The atomists tried to work out an atomic understanding of existence in general. For instance they constructed a cosmology using atoms that bears some resemblance to modern eternal inflationary cosmology in which there are infinitely many universes that are born out of primordial space-time foam and perhaps return to that—foam to foam. Unfortunately, the atomists got off on the wrong foot on the shape of the Earth: they were still flat Earthers when the round Earth theory was being established. Quite obviously to us, the atomists were badly non-experimental. Much of their thinking can be called rational myth. To a degree they were lucky in happening to be attracted to an essentially right idea. The atomists were eventually stigmatized as atheists: they did not deny that gods exist, but didn’t leave anything for the gods to do. This may have been their downfall. The more orthodox and popular philosophies of Plato, Aristotle, and the Stoics rejected atomism probably, among other things, for its seeming atheism. Christianity followed suit in this regard. The writings of the atomists only exist in fragments—and Democritos seems to have been as famous as Plato in his day. The Epicurean philosophers adopted atomism, but also suffered the stigmatization as atheists—and also hedonists who are, of course, the worst. But the atom idea lingered on through the centuries: Leucippos and Democritos, Epicurus, Lucretius (his surviving poem De Rerum Natura [On Nature] expounds atomism), Gassendi (17th century), Newton, Dalton: the chain is unbroken: it is not true that modern atomism has no historical or essential connection to ancient atomism. A good account of ancient atomism can be found in David Furley’s The Greek Cosmologists. Now, without recurring to the top of this preamble, atomism was invented in: a) the early 19th century. c) the 10th century CE.

d) the 5th century CE.

b) the 17th century by Gassendi. e) the 5th century BCE.

Chapt. 1 Classical Physics in Trouble 3 001 qmult 00800 1 1 1 easy memory: causality, relativity 8. Einstein ruled out faster than light signaling because: a) b) c) d) e)

it it it it it

would cause irresolvable causality paradoxes. would not cause irresolvable causality paradoxes. led to irresolvable paradoxes in quantum mechanics. would destroy the universe. had been experimentally verified.

001 qmult 00900 1 1 3 easy memory: EPR paradox 9. The Einstein-Podolsky-Rosen (EPR) paradox was proposed to show that ordinary quantum mechanics implied superluminal signaling and therefore was: a) b) c) d) e)

more or less correct. absolutely correct. defective. wrong in all its predictions. never wrong in its predictions.

001 qmult 01000 1 4 3 easy deducto-memory: Bell’s theorem 10. “Let’s play Jeopardy! For $100, the answer is: This theorem (if it is indeed inescapably correct) and the subsequent experiments on the effect the theorem dealt with show that quantum mechanical signaling exceeds the speed of light.” a) b) c) d) e)

What What What What What

is is is is is

Dark’s theorem, Alex? Midnight’s theorem, Alex? Bell’s theorem, Alex? Book’s theorem, Alex? Candle’s theorem, Alex?

Full-Answer Problems 001 qfull 00500 3 5 0 tough thinking: Rutherford’s nucleus Extra keywords: (HRW-977:62P) 1. Rutherford discovered the nucleus in 1911 by bombarding metal foils with alpha particles now known to be helium nuclei (atomic mass 4.0026). An alpha particle has positive charge 2e. He expected the alpha particles to pass right through the foils with only small deviations. Most did, but some scattered off a very large angles. Using a classical particle picture of the alpha particles and the entities they were scattering off of he came to the conclusion that atoms contained most of their mass and positive charge inside a region with a size scale of ∼ 10−15 m = 1 fm: this 10−5 times smaller than the atomic size. (Note fm stands officially for femtometer, but physicists call this unit a fermi.) Rutherford concluded that there must be a dense little core to an atom: the nucleus. a) Why did the alpha particles scatter off the nucleus, but not off the electrons? HINTS: Think dense core and diffuse cloud. What is the force causing the scattering? b) If the alpha particles have kinetic energy 7.5 Mev, what is their de Broglie wavelength? c) The closest approach of the alpha particles to the nucleus was of order 30 fm. Would the wave nature of the alpha particles have had any effect? Note the wave-particle duality was not even suspected for massive particles in 1911.

4

Chapt. 1 Classical Physics in Trouble

001 qfull 01000 3 5 0 tough thinking: black-body radiation, Wien law Extra keywords: (Le-62) gives a sketch of the derivations 2. Black-body radiation posed a considerable challenge to classical physics which it was partially able to meet. Let’s see how far we can get from a classical, or at least semi-classical, thermodynamic equilibrium analysis. a) Let Uλ be the radiation energy density per wavelength of a thermodynamic equilibrium radiation field trapped in some kind of cavity. The adjective thermodynamic equilibrium implies that the field is homogenous and isotropic. I think Hohlraum was the traditional name for such a cavity. Let’s call the field a photon gas and be done with it—anachronism be darned. Since the radiation field is isotropic, the specific intensity is then given by B(λ, T ) =

cUλ , 4π

(Pr.1)

where c is of course the speed of light. Specific intensity is radiation flux per wavelength per solid angle. From special relativity (although there may be some legitimately classical way of getting it), the momentum flux associated with a specific intensity is just B(λ, T )/c. Recall the rest plus kinetic energy of a particle is given by q E = p2 c2 + m20 c4 , (Pr.2) where p is momentum and m0 is rest mass. From an integral find the expression for the radiation pressure on a specularly reflecting surface: p=

1 U , 3

(Pr.3)

where p is now pressure and U is the wavelength-integrated radiation density. Argue that the same pressure applies even if the surface is only partially reflecting or pure blackbody provided the the radiation field and the surface are in thermodynamic equilibrium. HINT: Remember to account for angle of incidence and reflection. b) Now we can utilize a few classical thermodynamic results to show that U = aT 4 ,

(Pr.4)

where a is a radiation constant related to the Stefan-Boltzmann constant σ = 5.67051 × 105 ergs/(cm2 K4 ) and T is Kelvin temperature, of course. The relation between a and σ follows from the find the flux leaking out a small hole in the Hohlraum: F = 2π

Z

0

1

ca cU µ dµ = T 4 , 4π 4

(P r.5)

where µ is the cosine of the angle to the normal of the surface where the hole is. One sees that σ = ca/4. Classically a cannot be calculated theoretically; in quantum mechanical statistical mechanics a can be derived. The proportionality U ∝ T 4 can, however, be derived classically. Recall the 1st law of thermodynamics: dE = T dS − p dV , where E is internal energy, S is entropy, and V is volume. Note that     ∂E ∂E =T and = −p , ∂S V ∂V S

(Pr.6)

(Pr.7)

Chapt. 1 Classical Physics in Trouble 5 where the subscripts indicate the variables held constant. It follows from calculus (assuming well-behaved functions) that     ∂p ∂T =− , (Pr.8) ∂S V ∂V S The last relation is one of Maxwell’s four thermodynamic relations—Newton did things in threes; Maxwell in fours. Note that E = U V for a radiation field. Now go to it: show U ∝ T 4. c) As a by-product of the part (b) answer, you should have found that T ∝ V −1/3

(Pr.9)

for a quasistatic adiabatic process with the photon gas. (Find it now if somehow you missed it in the part (b) answer.) Assume you have a perfectly reflecting Hohlraum that you expand homologously by a scaling factor f (t), where t is time. Thus at any time t any length ℓ between physical points on the walls in the system is given by ℓ = f (t)ℓ0 ,

(Pr.10)

where ℓ0 was the physical length at t0 when f (t0 ) = 1. Find out how T , U , U dV , and E scale with f (t). What happens to the lost internal energy? HINT: This is easy. d) Consider the process described in the part (c) and show that λ = λ0 f (t)

(Pr.11)

for each specific intensity beam. Note you can use the non-relativistic Doppler effect since the velocity shift between scatterings off the walls is vanishingly small in the quasistatic limit. e) For the same system as in part (c) show that B(λ, T ) dλ dV = f (t)−1 B(λ0 , T0 ) dλ0 dV0 .

(Pr.12)

Then show that equation (Pr.12) leads naturally (if not absolutely necessarily so far as I can see) to the prescription for black-body specific intensity −5

B(λ, T ) = λ

 5 T g(x) = g(x) , x

(Pr.13)

where x ≡ λT

(Pr.14)

and g(x) is a universal function that cannot be determined from classical theory. Equation (Pr.13) is sometimes called Wien’s displacement law. However the name Wien’s displacement law is more usually (I think) reserved for the immediate result that for fixed T the the maximum of the black-body specific intensity (i.e., the maximum of x−5 g(x)) occurs at a wavelength given by xmax λ= , (Pr.15) T where xmax is the global universal location of maximum for the universal function g(x). It was empirically known that black-body radiation had only one maximum with wavelength, and so this corresponds to xmax . I think classically xmax has to be determined empirically. Wien’s radiation law was I believe a fit to the observations of Wien’s displacement law. This law is   5  k2 T , (Pr.16) exp − B(λ, T ) = k1 x x

6

Chapt. 1 Classical Physics in Trouble where k1 and k2 had to be determined from the fit. Wien’s law works well for short wavelengths > > xmax ), but failed badly at shorter wavelengths (Pa-190, but note the x there is the inverse of the x here aside from a constant). In fact the Rayleigh-Jeans law goes to inifinity as x goes to zero and the total energy in a Rayleigh-Jeans radiation field is infinite (Le-64): this is sometimes called the ultraviolet catastrophe (BFG-106). The correct black-body specific intensity law was derived from a primitive quantum theory by Max Planck in 1900 (BFG-106). Planck obtained an empirically excellent fit to the black-body specific intensity and then was able to derive it from his quantum hypothesis. The Rayleigh-Jeans and Planck laws are the subject for another question.

001 qfull 01100 2 5 0 moderate thinking: Bohr atom 3. In 1913, Niels Bohr presented his model of the hydrogen atom which was quickly generalized to the hydrogenic atom (i.e., the one-electron atom of any nuclear charge Z). This model correctly gives the main hydrogenic atom energy levels and consists of a mixture of quantum mechanical and classical ideas. It is historically important for showing that quantization is somehow important in atomic structure and pedagogically it is of interest since it shows how simple theorizing can be done. But the model is, in fact, incorrect and from the modern perspective probably even misleading about the quantum mechanical nature of the atom. It is partially an accident of nature that it exists to be found. Only partially an accident since it does contain correct ingredients. And it is no accident that Bohr found it. Bohr knew what he wanted: a model that would successfully predict the hydrogen atom spectrum which is a line spectrum showing emission at fixed frequencies. He knew from Einstein’s photoelectric effect theory that electromagnetic radiation energy was quantized in amounts hν where h = 6.62606896(33) × 10−27 erg s was Planck’s constant (which was introduced along with the quantization notion to explain blackbody radiation in 1900) and ν was frequency of the quantum of radiation. He recognized that Planck’s constant had units of angular momentum. He knew from Rutherford’s nuclear model of the atom that the positive charge of an atom was concentrated in region that was much smaller than the atom size and that almost all the mass of the atom was in the nucleus. He knew that there were negative electrons in atoms and they were much less massive than the nucleus. He knew the structure of atoms was stable somehow. By a judicious mixture of classical electromagnetism, classical dynamics, and quantum ideas he found his model. A more sophisticated mixture of these concepts would lead to modern quantum mechanics. Let’s see if we can follow the steps of the ideal Bohr—not the Bohr of history. NOTE: This a semi-classical question: Bohr, ideal or otherwise, knew nothing of the Schr¨odinger equation in 1913. Also note that this question uses Gaussian CGS units not MKS units. The most relevant distinction is that electric charge eMKS eCGS = √ 4πǫ0 which implies the fine structure constant in CGS is e2 α= − . hc Astronomy is all Gaussian CGS by the way. a) Bohr thought to build the electron system about the nucleus based on the electrostatic inverse square law with the electron system supported against collapse onto the nucleus by angular momentum. The nucleus was known to be much tinnier than the electron system

Chapt. 1 Classical Physics in Trouble 7 which gives the atom its volume. The nucleus could thus be a considered an immobile point center of force at the origin of the relative nucleus-electron coordinate system frame. This frame is non-inertial, but classically can be given an inertial-frame treatment if the electron is given a reduced mass given by m=

  me mnucleus me , ≈ me 1 − me + mnucleus mnucleus

where me the electron mass and mnucleus is the nucleus mass. The approximation is valid for me /mnucleus Vmin for any normalizable ¯ = hHi = Vmin never holds quantum state of the system considered. Note the equality E mechanically. (There is an over-idealized exception, which we consider in part (e).) Prove the inequality. HINTS: The key point is to show that hT i > 0 for all physically allowed states. Use integration by parts. ¯ > Vmin implies E > Vmin , where E is any eigen-energy of the c) Now show that result E system considered. Note the equality E = Vmin never holds quantum mechanically (except for the over-idealized system considered in part (e)). In a sense, there is no rest state for quantum mechanical particle. This lowest energy is called the zero-point energy. d) The E > Vmin result for an eigen-energy in turn implies a 3rd result: any ideal measurement always yields an energy greater than Vmin Prove this by reference to a quantum mechanical postulate. e) This part is NOT to be done on EXAMS: it’s just too much (for the grader). There is actually an exception to E > Vmin result for an eigen-energy where E = Vmin occurs. The exception is for quantum mechanical systems with periodic boundary conditions and a constant potential. In ordinary 3-dimensional Euclidean space, the periodic boundary conditions can only occur for rings (1-dimensional systems) and sphere surfaces (2dimensional systems) I believe. Since any real system must have a finite size in all 3 spatial dimensions, one cannot have real systems with only periodic boundary conditions.

22 Chapt. 2 QM Postulates, Schr¨odinger Equation, and the Wave Function Thus, the exception to the E > Vmin result is for unrealistic over-idealized systems. Let us consider the idealized ring system as an example case. The Hamiltonian for a 1-dimensional ring with a constant potential is H=−

2 − h ∂2 +V , 2mr2 ∂φ2

where r is the ring radius, φ is the azimuthal angle, and V is the constant potential. Find the eigen-functions and eigen-energies for the Schr¨odinger equation for the ring system with periodic boundary conditions imposed. Why must one impose periodic boundary conditions on the solutions? What solution has eigen-energy E = Vmin ? 002 qfull 00110 2 5 0 moderate thinking: beyond the classical turning points 14. The constant energy of a classical particle in a conservative system is given by E =T +V . Since classically T ≥ 0 always, a bound particle is confined by surface defined by T = 0 or E = V (~r ). The points constituting this surface are called the turning points: a name which makes most sense in one dimension. Except for static cases where the turning point is trivially the rest point (and maybe some other weird cases), the particle comes to rest only for an instant at a turning point since the forces are unbalanced there. So it’s a place where a particle “ponders for an instant before deciding where to go next”. The region with V > E is classically forbidden. Now for most quantum mechanical potential wells, the wave function extends beyond the classical turning point surface into the classical forbidden zone and, in fact, usually goes to zero only at infinity. If the potential becomes infinite somewhere (which is an idealization of course), the wave function goes to zero: this happens for the infinite square well for instance. Let’s write the 1-dimensional time-independent Schr¨odinger equation in the form 2m ∂2ψ = 2 (V − E)ψ . 2 − ∂x h a) Now solve for ψ for the region with V > E with simplifying the assumption that V is constant in this region. b) Can the solutions be normalized? c) Can the solutions constitute an entire wave function? Can they be part of a wave function? In which regions? d) Although we assumed constant V , what crudely is the behavior of the wave function likely to be like the regions with V > E. e) For typical potentials considered at our level, qualitatively what is the likelihood of finding the particle in the classically forbidden region? Why? 002 qfull 01100 3 5 0 tough thinking: 1-d non-degeneracy 15. If there are no internal degrees of freedom (e.g., spin) and they are NORMALIZABLE, then one-particle, 1-dimensional energy eigenstates are non-degenerate. We (that is to say you) will prove this. Actually, we know already that any 2nd order ordinary linear differential equation has only two linearly independent solutions (Ar-402) which means, in fact, that from the start we know there is a degeneracy of 2 at most. Degeneracy count is the number of independent solutions. If there is more than one independent solution, then infinitely many linear combinations of solutions have the same energy. But in an expansion of wave function, only a set linear independent solutions is needed and thus the number of such solutions is the physically relevant

Chapt. 2 QM Postulates, Schr¨odinger Equation, and the Wave Function 23 degeneracy. Of course, our proof means that one of the linearly independent solutions is not normalizable. a) Assume you have two degenerate 1-dimensional energy eigenstates for Hamiltonian H: ψ1 and ψ2 . Prove that ψ1 ψ2′ − ψ2 ψ1′ equals a constant where the primes indicate derivative with respect to x the spatial variable. HINT: Write down the eigenproblem for both ψ1 and ψ2 and do some multiplying and subtraction and integration. b) Prove that the constant in part (a) result must be zero. HINT: To be physically allowable eigenstates, the eigenstates must be normalizable. c) Integrate the result of the part (b) answer and show that the two assumed solutions are not physically distinct. Show for all x that ψ2 (x) = Cψ1 (x) , where C is a constant. This completes the proof of non-degeneracy since eigenstates that differ by a multiplicative constant are not physically (i.e., expansion) distinct. HINT: You have to show that there is no other way than having ψ2 (x) = Cψ1 (x) to satisfy the condition found in the part (b) answer. Remember the eigenproblem is a linear, homogeneous differential equation. 002 qfull 01200 2 3 0 mod math: 3-d exponential wave function, probability Extra keywords: (Co1-342:6), 3-d wave function, probability, momentum representation 16. Consider the 3-dimensional wave function # " X |xi |/(2ai ) , Ψ(~r ) = A exp − i

where the sum runs over the three Cartesian coordinates and the ai ’s are real positive length parameters. a) Calculate the normalization factor A. HINT: Recall that the integrand is |Ψ(~r )|2 = Ψ(~r )∗ Ψ(~r ). I’m always forgetting this myself when the function is pure real and there is no imaginary part to remind me of it. b) Calculate the probability that a measurement of xi will yield a result between 0 and ai , where i could be any of the three coordinates. HINT: There are no restrictions on values of the other coordinates: they could be anything at all. Thus one just integrates over all of those other coordinate positions remembering normalization of course. c) Calculate the probability that simultaneous measurements of xj and xk will yield results in the ranges −aj to aj and −ak to ak , respectively. The j and k could be any pair of the two coordinates. HINT: Remember the hint for part (b). d) Calculate the probability that a measurment of momentum will yield a result in the element dpi dpj dpk centered at the point pi = pj = 0, pk = − h /ak . HINT: You will need to find the momentum representation of the state.

Chapt. 3 Infinite Square Wells and Other Wells

Multiple-Choice Problems 003 qmult 00050 1 1 1 easy memory: infinite square well 1. In quantum mechanics, the infinite square well can be regarded as the prototype of: a) all bound systems. b) all unbound systems. c) both bound and unbound systems. d) neither bound nor unbound systems. e) Prometheus unbound. 003 qmult 00100 2 4 2 moderate deducto-memory: infinite square well BCs 2. In the infinite square well problem, the wave function and its first spatial derivative are: a) b) c) d) e)

both continuous at the boundaries. continuous and discontinuous at the boundaries, respectively. both discontinuous at the boundaries. discontinuous and continuous at the boundaries, respectively. both infinite at the boundaries.

003 qmult 00300 1 1 3 easy memory: boundary conditions 3. Meeting the boundary conditions of bound quantum mechanical systems imposes: a) Heisenberg’s uncertainty principle. b) Schr¨odinger’s equation. d) a vector potential. e) a time-dependent potential.

c) quantization.

003 qmult 00400 1 1 5 easy memory: continuum of unbound states 4. At energies higher than the bound stationary states there: a) are between one and several tens of unbound states. b) are only two unbound states. c) is a single unbound state. e) is a continuum of unbound states.

d) are no states.

003 qmult 00500 1 4 2 easy deducto-memory: tunneling 5. “Let’s play Jeopardy! For $100, the answer is: This effect occurs because wave functions can extend (in an exponentially decreasing way albeit) into the classically forbidden region: i.e., the region where a classical particle would have negative kinetic energy.” What is

, Alex?

a) stimulated radiative emission b) quantum mechanical tunneling d) symmetrization e) normalization

c) quantization

003 qmult 00600 2 1 2 moderate memory: benzene ring model 6. A simple model of the outer electronic structure of a benzene molecule is a 1-dimensional infinite square well with: a) vanishing boundary conditions. c) aperiodic boundary conditions. e) incorrect boundary conditions.

b) periodic boundary conditions. d) no boundary conditions.

24

Chapt. 3 Infinite Square Wells and Other Wells

25

Full-Answer Problems 003 qfull 00100 2 3 0 moderate math: infinite square well in 1-d 1. You are given the time-independent Schr¨odinger equation # 2 − h ∂2 Hψ(x) = − + V (x) ψ(x) = Eψ(x) 2m ∂x2 "

and the infinite square well potential V (x) =

n

0 , x ∈ [0, a]; ∞ otherwise.

a) What must the wave function be outside of the well (i.e., outside of the region [0, a]) in order to satisfy the Schr¨odinger equation? Why? b) What boundary conditions must the wave function satisfy? Why must it satisfy these boundary conditions? c) Reduce Schr¨odinger’s equation inside the well to an equation of the same form as the CLASSICAL simple harmonic oscillator differential equation with all the constants combined into a factor of −k 2 , where k is newly defined constant. What is k’s definition? d) Solve for the general solution for a SINGLE k value, but don’t impose boundary conditions or normalization yet. A solution by inspection is adequate. Why can’t we allow solutions with E ≤ 0? Think carefully: it’s not because k is imaginary when E < 0. e) Use the boundary conditions to eliminate most of the solutions with E > 0 and to impose quantization on the allowed set of distinct solutions (i.e., on the allowed k values). Give the general wave function with the boundary conditions imposed and give the quantization rule for k in terms of a dimensionless quantum number n. Note that the multiplication of a wave function by an arbitrary global phase factor eiφ (where φ is arbitrary) does not create a physically distinct wave function (i.e., does not create a new wave function as recognized by nature.) (Note the orthogonality relation used in expanding general functions in eigenfunctions also does not distinguish eigenfunctions that differ by global phase factors either: i.e., it gives the expansion coefficients only for distinct eigenfunctions. So the idea of distinct eigenfunctions arises in pure mathematics as well as in physics.) f) Normalize the solutions. g) Determine the general formula for the eigenenergies in terms of the quantum number n. 003 qfull 00400 2 3 0 moderate math: moments of infinite square well Extra keywords: (Gr-29:2.4) 2. Calculate hxi, hx2 i, hpi, hp2 i, σx , and σp for the 1-dimensional infinite square well with range [0, a]. Recall the general solution is ψ=

r

2 sin(kx) = a

r

 nπ  2 sin x , a a

where n = 1, 2, 3, . . . . Also check that the Heisenberg uncertainty principle is satisfied. 003 qfull 00450 2 3 0 moderate math: infinite square well features

26 Chapt. 3 Infinite Square Wells and Other Wells 3. The one-dimensional infinite square well with a symmetric potential and width a is  0 for |x| ≤ a/2; V = ∞ for |x| > a/2. The eigenstates for infinite square well are given by r  2 cos(kx) for n = 1, 3, 5 . . .; ψn (x) = × sin(kx) for n = 2, 4, 6 . . ., a where

ka nπ nπ = and k= . 2 2 a The n is the quantum number for eigenstates. The eigenstates have been normalized and are guaranteed orthogonal by the mathematics of Hermitian operators of the which the Hamiltonian is one. A quantum number is a dimensionless index (usually integer or half-integer) that specifies the eigenstates and eigenvalues somehow. The eigen-energies are given by En =

2 2 − − h  π 2 2 h k2 n . = 2m 2m a

a) Verify the normalization of eigenstates. b) Determine hxi for the eigenstates. c) Determine hpop i for the eigenstates. HINT: Recall pop =

− h ∂ . i ∂x

d) Determine hp2op i and the momentum standard deviation σp for the eigenstates.

e) Determine hx2 i and the position standard deviation σx in the large n limit. HINT: Assume x2 can be approximated constant over one complete cycle of the probability density ψn∗ ψn f) Now for the boring part. Determine hx2 i and the position standard deviation σx exactly now. HINT: There probably are several different ways of doing this, but there seem to be no quick tricks to the answer. The indefinite integral Z 2 2 x2 sin(bx) + 2 x cos(bx) − 3 sin(bx) x2 cos(bx) dx = b b b might be helpful.

g) Verify that the Heisenberg uncertainty principle ∆x∆p = σx σp ≥

− h 2

is satisfied for the infinite square well case. 003 qfull 00500 3 5 0 tough thinking: mixed infinite square well stationary states Extra keywords: (Gr-29:2.6) 4. A particle is in a mixed state in a 1-dimensional infinite square well where the well spans [0, a] and the solutions are in the standard form of Gr-26. At time zero the state is Ψ(x, 0) = A [ψ1 (x) + ψ2 (x)] ,

Chapt. 3 Infinite Square Wells and Other Wells

27

where ψ1 (x) and ψ2 (x) are the time-independent 1st and 2nd stationary states of the infinite square well. a) Determine the normalization constant A. Remember the stationary states are orthonormal. Also is the normalization a constant with time? Prove this from the general time evolution equation   1 ∂Q dhQi + − hi[H, Q]i . = dt ∂t h b) Now write down Ψ(x, t). Give the argument for why it is the solution. As a simplfication in the solution use − h  π 2 E1 ω1 = − = , 2m a h where E1 is the ground state energy of the infinite square well. c) Write out |Ψ(x, t)|2 and simplify it so that it is clear that it is pure real. Make use Euler’s formula: eix = cos x + i sin x. What’s different about our mixed state from a stationary state? d) Determine hxi for the mixed state. Note that the solution is oscillatory. What is the angular frequency wq and amplitude of the oscillation. Why would you be wrong if your amplitude was greater than a/2. e) Determine hpi for the mixed state. As Peter Lorre (playing Dr. Einstein—Herman Einstein, Heidelberg 1919) said in Arsenic and Old Lace “the quick way, Chonny.” f) Determine hHi for the mixed state. How does it compare to E1 and E2 ? g) Say a classical particle had kinetic energy equal to the energy hHi found in the part (f) answer. The particle is bounces back and forth between the walls of the infinite square well. What would its angular frequency be in terms of ωq and the angular frequency found in the part (d) answer. 003 qfull 00600 2 5 0 moderate thinking: revival time Extra keywords: Gr-85 The hints make it possible as new test problem. 5. The revival time is the minimum time period for a wave function to repeat (i.e., to cycle back to its original form), or slightly less restrictively, for the probability density to repeat. a) Say we had a system with eigen-energies given by the formula En = E1 f (n) + E0 , where E0 is a zero-point energy, n is a quantum number that runs 1, 2, 3, . . . or 0, 1, 2, 3, . . . , f (n) is a strictly increasing function that always has an integer value, and f (1) = 1. What is the revival time (in the probability density sense) for general wave function Ψ(x, t) for this system? HINT: The zero-point energy gives a time-dependent global phase factor for any expansion in the stationary states, and thus cancels out of the probability density. Assume orthonormal energy-eigen states and recall X Ψ(x, t) = cn e−iωn t ψn (x) , n

where ωn = En /− h are the angular frequencies and ψn (x) are the energy-eigenstates. The period for ωn is 2π/ωn .

b) The eigen-energies for the infinite square well and the simple harmonic oscillator are, respectively,   2 − h  π 2 2 1 − h ωcl , En = n and En = n + 2m a 2

28 Chapt. 3 Infinite Square Wells and Other Wells where n = 1, 2, 3, . . . for the infintie square well and 0, 1, 2, 3, . . . for the simple harmonic oscillator, m is the particle mass, a is the well width, and ωcl simple harmonic oscillator frequency which enters the quantum mechanical description as parameter in the simple harmonic oscillator potential. What are the revival times for general wave functions for these two systems? c) What are the classical revival times for a particle in a infinite square well system and in a simple harmonic oscillator system in terms of, respectively, energy and ωcl ? The classical times are just the oscillation periods for the particles. The particle in the infinite square well is assumed to be just bouncing between the walls without loss of kinetic energy. d) For what classical energy E in units of E1 (i.e., E/E1 ) are the quantum mechanical and classical revival times equal for the infinite square well? What is the relationship between the quantum mechanical and classical revival times for the simple harmonic oscillator? 003 qfull 01000 3 5 0 tough thinking: 3-d infinite cubical well Extra keywords: (Gr-124:4.2), separation of Schr¨odinger equation 6. Consider an infinite cubical well or particle-in-a-box system. The potential is V (x, y, x) =



0, for x, y, and z in the range 0 to a; ∞, otherwise.

The wave functions must be zero at the boundaries for an infinite well recall. a) Solve for the stationary states from the 3-dimensional Schr¨odinger equation and find their energies in terms of quantum numbers nx , ny , and nz . HINTS: Separate the Schr¨odinger equation into x, y, and z components. Identify the sum of the separation constants as energy or, if you prefer, energy times a constant. Solve separately matching the boundary conditions and then assemble the normalized TOTAL SOLUTION. Of course, all three dimensions behave the same so only one of them really needs to be done—which is NOT to say that each one is a total solution all by itself. b) Is there energy degeneracy? Why? c) Determine the 6 lowest energies and their degeneracy? HINTS: A systematic approach would be fix an nmax = max(nx , ny , nz ) and count all energies and their degeneracies governed by that nmax . One works one’s way up from nmax = 1 to as high as one needs to go to encompass the 6 lowest energies. Each nmax governs the energies between n2max + 2 and 3n2max (where we have written the in dimensionless form). Note, e.g., that states described by (nx = 4, ny = 1, nz = 1), (nx = 1, ny = 4, nz = 1), and (nx = 1, ny = 1, nz = 4) are all distinct and degenerate. 003 qfull 01100 1 3 0 easy math: pi-states of a benzene ring Extra keywords: (Ha-323:2.1) 7. Imagine that we have 6 free electrons in 1-d circular system of radius r = 1.53 ˚ A. This system is a simple model of a benzene ring molecule (C6 H6 ) of 6 carbon atoms each bonded to a hydrogen (Ke-153). The carbons are bonded by bonded by a single-double bond superposition. The free electron system on the benzene constitute the benzene pi-states. a) Obtain expressions for the eigenstates, wavenumbers, and eigen-energies of the free electrons. Re-express the wavenumbers and energies in terms of Angstroms and 2 electronvolts. Note − h /(2m) = 3.81 eV-˚ A2 for electrons. Sketch the energy level diagram. b) One electron per carbon lies in the circular state for a benzene ring: these are the π electrons. Assuming that two electrons can be found in any state, what is the total energy of the ground state configurations? NOTE: Two electrons can be found in any state because there are two spin states they can be found in. Thus the Pauli exclusion principle is maintained: i.e., only one electron can be found in any single-particle state (e.g., Gr-180).

Chapt. 3 Infinite Square Wells and Other Wells

29

c) What is the energy difference in eV between the lowest empty level and highest occupied level for the ground state configuration? This is the radiation absorption threshold. What is the threshold line wavelength in microns? In what wavelength regime is this line? NOTE: The constant hc = 1.23984 eV-µm. d) Now imagine we broke the benzene ring, but magically kept the length constant. Obtain expressions for the eigenstates, wavenumbers, and eigen-energies of the free electrons. Reexpress the wavenumbers and energies in terms of Angstroms and electronvolts. Sketch the energy levels on the previous energy level diagram. e) What is the ground state energy for the broken ring. What is the change in ground state energy from the unbroken ring. This change is a contribution to the energy required to break the ring or the energy of a resonant π bond. f) I know we said that somewhere that quantum mechanical bound states always had to have E > Vmin . But in the ring case we had Vmin = 0, and we have a state with E = 0. So why do we have this paradox? Is the paradox possible in 2 dimensions or 3 dimensions?

Chapt. 4 The Simple Harmonic Oscillator (SHO)

Multiple-Choice Problems 004 qmult 00100 2 4 1 moderate deducto-memory: SHO eigen-energies 1. “Let’s play Jeopardy! For $100, the answer is: − h ω. a) What is the energy difference between adjacent simple harmonic ocsillator energy levels, Alex? b) What is the energy difference between adjacent infinite square well energy levels, Alex? c) What is the energy difference between most adjacent infinite square well energy levels, Alex? d) What is the energy difference between the first two simple harmonic ocsillator energy levels ONLY, Alex? e) What is the bar where physicists hang out in Las Vegas, Alex?

Full-Answer Problems 004 qfull 00100 2 3 0 moderate math: SHO ground state analyzed Extra keywords: (Gr-19:1.14) 1. The simple harmonic oscillator (SHO) ground state is Ψ0 (x, t) = Ae−β where

− hω E0 = 2

and

2

x2 /2−iE0 t/− h

β=

r

,

mω . − h

a) Verify that the wave function satisfies the full Schr¨odinger equation for the SHO. Recall that the SHO potential is V (x) = (1/2)mω 2 x2 . b) Determine the normalization constant A. c) Calculate the expectation values of x, x2 , p, and p2 . d) Calculate σx and σp , and show that they satisfy the Heisenberg uncertainty principle. 004 qfull 00200 2 3 0 moderate thinking: SHO classically forbidden Extra keywords: (Gr-43:2.15) classical turning points 2. What is the probability is of finding a particle in the ground state of a simple harmonic oscillator potential outside of the classically allowed region: i.e., beyond the classical turning points? HINT: You will have to use a table of the integrated Gaussian function. 004 qfull 00300 2 5 0 moderate thinking: mixed SHO stationary states Extra keywords: (Gr-43:2.17) 30

Chapt. 4 The Simple Harmonic Oscillator (SHO)

31

3. A particle in a simple harmonic oscillator (SHO) potential has initial wave function Ψ(x, 0) = A [ψ0 + ψ1 ] , where A is the normalization constant and the ψi are the standard form 0th and 1st SHO eigenstates. Recall the potential is V (x) =

1 mω 2 x2 . 2

Note ω is just an angular frequency parameter of the potential and not NECESSARILY the frequency of anything in particular. In the classical oscillator case ω is the frequency of oscillation, of course. a) Determine A assuming it is pure real as we are always free to do. b) Write down Ψ(x, t). There is no need to express the ψi explicitly. Why must this Ψ(x, t) be the solution? c) Determine |Ψ(x, t)|2 in simplified form. There should be a sinusoidal function of time in your simplified form. d) Determine hxi. amplitude.

Note that hxi oscillates in time. What is its angular frequency and

e) Determine hpi the quick way using the 1st formula of Ehrenfest’s theorem. Check that the 2nd formula of Ehrenfest’s theorem holds. 004 qfull 01000 3 5 0 tough thinking: infinite square well/SHO hybrid Extra keywords: (Mo-424:9.4) 4. Say you have the potential (∞ , x < 0; V (x) = 1 mω 2 x2 x ≥ 0. 2 a) By reflecting on the nature of the potential AND on the boundary conditions, identify the set of Schr¨odinger equation eigenfunctions satisfy this potential. Justify your answer. HINTS: Don’t try solving the Schr¨odinger equation directly, just use an already known set of eigenfunctions to identify the new set. This shouldn’t take long. b) What is the expression for the eigen-energies of your eigenfunctions? c) What factor must multiply the already-known (and already normalized) eigenfunctions you used to construct the new set you found in part (a) in order to normalize the new eigenfunctions? HINT: Use the evenness or oddness (i.e., definite parity) of the alreadyknown set. d) Show that your new eigenfunctions are orthogonal. HINT: Use orthogonality and the definite parity of the already-known set. e) Show that your eigenfunctions form a complete set given that the already-known set was complete. HINTS: Remember completeness only requires that you can expand any suitably well-behaved function (which means I think it has to be piecewise continuous (Ar435) and square-integrable (CDL-99) satisfying the same boundary conditions as the set used in the expansion. You don’t have to be able to expand any function. Also, use the completeness of the already-known set. 004 qfull 01100 3 5 0 tough thinking: Hermite polynomials 1

32 Chapt. 4 The Simple Harmonic Oscillator (SHO) 5. The generating function method is a powerful method for obtaining the eigenfunctions of SturmLiouville Hermitian operators and some of their general properties. One can possibly obtain with only moderately arduous labor some special values, the norm value, a general series formula for the eigenfunctions, and recurrence relations for iteratively constructing the complete set of eigenfunctions. The only problem is who the devil thought up the generating function? In the case of Hermite polynomials, the generating function—which may or may not have been thought up by French mathematician Charles Hermite (1822–1901)—is 2

g(x, t) = e−t

+2tx

=

∞ X

Hn

n=0

tn n!

(Ar-609ff; WA-644). The Hn are the Hermite polynomials: they are functions of x and n is their order. 2 Actually, the HERMITE EQUATION needs a weight function e−x to be put in SturmLiouville self-adjoint form (Ar-426, WA-486). Alternatively, the Hermite polynomials times 2 e−x /2 satisfy a Sturm-Liouville Hermitian operator equation which happens to be the timeindependent Schr¨odinger equation for the 1-dimensional quantum mechanical simple harmonic oscillator (Ar-612, WA-638). The 1-dimensional quantum mechanical simple harmonic oscillator is one of those few quantum mechanical systems with an analytic solution. NOTE: The parts of this question are independent: i.e., you should be able to do any of the parts without having done the other parts. a) Find the 1st recurrence relation Hn+1 = 2xHn − 2nHn−1 by differentiating both the generating function and its and series expansion with respect to t. This recurrence relation provides a means of finding any order of Hermite polynomial. HINT: You will need to re-index summations and make use of the uniqueness theorem of power series. b) Find the 2nd recurrence relation Hn′ = 2nHn−1 by differentiating both the generating function and its and series expansion with respect to x. HINT: You will need to re-index summations and make use of the uniqueness theorem of power series. c) Use the 1st recurrence relation to work out and tabulate the polynomials up to 3rd order: i.e., find H0 , H1 , H2 , and H3 . You can find the first two polynomials (i.e., the 0th and 1st order polynomials) needed to start the recurrence process by a simple Taylor’s series expansion of generating function. d) Use the 1st recurrence relation to prove that the order of a Hermite polynomial agrees with its polynomial degree (which is the degree of its highest degree term) and that even order Hermite polynomials are even functions and the odd order ones are odd functions. The last result means that the Hermite polynomials have definite parity (i.e., are either even or odd functions). HINT: Use proof by induction and refer to collectively to the results to be proven as “the results to be proven”. If you didn’t get H0 and H1 explicitly in part (c), you can assume H0 has degree 0 and H1 has degree 1. 004 qfull 01110 3 5 0 tough thinking: Hermite polynomials 2 6. Now for some more Hermite polynomial results. Recall the Hermite polynomial generating function is ∞ X 2 tn g(x, t) = e−t +2tx = Hn n! n=0

Chapt. 4 The Simple Harmonic Oscillator (SHO)

33

(Ar-609ff; WA-644). The Hn are the Hermite polynomials: they are functions of x and n is their order. Also recall the two recurrence relations Hn+1 = 2xHn − 2nHn−1

Hn′ = 2nHn−1

and

and the first few Hermite polynomials which are given in the table below. Table: Hermite Polynomials Order

Polynomial

0 1 2 3 4 5 6

H0 H1 H2 H3 H4 H5 H6

=1 = 2x = 4x2 − 2 = 8x3 − 12x = 16x4 − 48x2 + 12 = 32x5 − 160x3 + 120x = 64x6 − 480x4 + 720x2 − 120

NOTE: The parts of this question are largely independent: i.e., you should be able to do most parts without having done others. a) Now for something challenging. Show that

g(x, t) = e

−t2 +2tx

=

∞ X (−t2 + 2tx)ℓ

ℓ!

ℓ=0

∞ n [n/2] X t X n! = (−1)k (2x)n−2k n! (n − 2k)!k! n=0 k=0

which implies that [n/2]

Hn =

n! (−1)k (2x)n−2k . (n − 2k)!k!

X

k=0

Note [n/2] =



n/2 for n even; (n − 1)/2 for n odd.

HINTS: You will have to expand (−t2 + 2tx)n in a binomial series and then re-order the summation. A schematic table of the terms ordered in row by ℓ and in column by k makes the re-ordering of the summation clearer. One should add up diagonals rather than rows. b) Prove the following special results from the generating function: H2n (0) = (−1)n

(2n)! , n!

Hn (x) = (−1)n Hn (−x) .

H2n+1 (0) = 0 ,

The last results shows that the Hermite polynomials have definite parity: even for n even; odd for n odd. c) What is called the Rodrigues’s formula for the Hermite polynomials can also be derived from the generating function: Hn = (−1)n ex

2

∂ n −x2 (e ). ∂xn

Derive this formula. HINTS: Write 2

g(x, t) = e−t

+2tx

2

= ex e−(t−x)

2

34 Chapt. 4 The Simple Harmonic Oscillator (SHO) and note that

∂f (t − x) ∂f (t − x) =− . ∂t ∂x

d) Now show that Hermite’s differential equation Hn′′ − 2xHn′ + 2nHn = 0 follows from the two recurrence relations. This result shows that the Hermite polynomials satisfy Hermite’s differential equation. e) Now consider the Hermite differential equation h′′ − 2xh′ + 2νh = 0 , where ν is not necessarily an integer ≥ 0. Try a power series solution h=

∞ X

aℓ xℓ ,

ℓ=0

and show for sufficiently large ℓ and x that the series solutions approximate growing 2 2 exponentials of the form ex and xex —unless ν is a positive or zero integer in which case one gets what kind of solution? f) Hey this question is just going on and on. The Hermite differential equation cannot be written in an eigenproblem form with a Hermitian operator since the operator ∂ ∂2 − 2x ∂x2 ∂x is not, in fact, Hermitian. I won’t ask you to prove this since I don’t what to do that myself tonight. But if you substitute for Hn (x) (with n a positive or zero integer) the function ψn (x)ex

2

/2

in the Hermite differential equation, you do an eigenproblem with a Hermitian operator. Find this eigenproblem equation. What are the eigenfunctions and eigenvalues? Are the eigenfunctions square-integrable: i.e., normalizable in a wave function sense? Do the eigenfunctions have definite parity? Are the eigenvalues degenerate for square-integrable solutions? Based on a property of eigenfunctions of a Hermitian operator what can you say about the orthogonality of the eigenfunctions? g) In order to normalize the eigenfunctions of part (i) in a wave function sense consider the relation ∞ X 2 2 2 2 sm tn −x2 e Hm Hn = e−x g(x, s)g(x, t) = e−x e−s +2sx e−t +2tx . m! n! m,n=0

Integrate both sides over all x and use uniqueness of power series to find the normalization constants and incidently verify orthogonality. h) Now here in part infinity we will make the connection to physics. The simple harmonic oscillator time independent Schr¨odinger equation is 2 − h ∂2ψ 1 + mω 2 y 2 ψ = Eψ . − 2m ∂y 2 2

Chapt. 4 The Simple Harmonic Oscillator (SHO)

35

One can reduce this to the dimensionless eigenproblem of part (i), by changing the variable with x = βy . To find β, let

2 − 1 h and B = mω 2 2m 2 and divide equation through by an unknown C, equate what needs to be equated, and solve for C and β. What are the physical solutions and eigen-energies of the SHO eigenproblem?

A=

i) Check to see if Charles Hermite (1822–1901) did think up the Hermite polynomial generating function at the St. Andrews MacTutor History of Mathematics Archive: http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html HINTS: You don’t have to do this in a test mise en sc`ene.

Chapt. 5 Free Particles and Momemtum Representation

Multiple-Choice Problems 005 qmult 00100 1 1 2 easy deducto-memory: definition free particle 1. A free particle is: a) bound. b) unbound. c) both bound and unbound. d) neither bound nor unbound. e) neither here nor there. 005 qmult 00200 1 4 5 easy deducto-mem: free particle system 2. The free particle system is one with where the potential is: a) the simple harmonic oscillator potential (SHO). b) a quasi-SHO potential. c) an infinite square well potential. d) a finite square well potential. e) zero (or a constant) everywhere. 005 qmult 00300 1 4 4 easy deducto-mem: free particle eigenfunction 3. The general expression for the free particle energy eigenfunction in 1-dimension is: q 2 ikx kx kx h . a) e , where k = ±E. b) e , where k = ±E. c) e , where k = ± 2mE/− q q 2 2 2 h . e) ekx , where k = ± 2mE/− h . d) eikx , where k = ± 2mE/−

005 qmult 00400 1 4 1 easy deducto-mem: free particle normalization 1 4. The free particle energy eigenfunctions are not physical states that a particle can actually be in because they: a) b) c) d) e)

can’t be normalized (i.e., they arn’t square-integrable). can be normalized (i.e., they are square-integrable). are growing exponentials. don’t exist. do exist.

005 qmult 00500 1 1 3 easy memory: free particle normalization 2 5. The free particle stationary states: a) can be occupied by a particle. b) can be occupied by two particles. c) cannot actually be occupied by a particle. d) are unknown. e) are normalizable.

Full-Answer Problems 005 qfull 00100 2 5 0 easy thinking: momentum representation Extra keywords: (Gr-49:2.21) 1. The initial wave function of a free particle is  A , x ∈ [−a, a]; Ψ(x, 0) = 0 , otherwise, 36

Chapt. 5 Free Particles and Momemtum Representation 37 where a and A are positive real numbers. The particle is in a completely zero potential environment since it is a free particle. a) Determine A from normalization. b) Determine ψ(k) = Ψ(k, 0) the time-zero wavenumber representation of the particle state. It is the Fourier transform of Ψ(x, 0). What is Ψ(k, t)? Sketch ψ(k). Locate the global maximum and the zeros of ψ(k). Give the expression for the zeros (i.e., for the location of the zeros). c) Determine the wavenumber space probability density |Ψ(k, t)|2 and show then that Ψ(k, t) is normalized in wavenumber space. (You can use a table integral.) Sketch |Ψ(k, t)|2 and locate the global maximum and the zeros. Give the expression for the zeros. d) Crudely estimate and then calculate exactly σx , σk , and σp for time zero. Are the results consistent with the Heisenberg uncertainty principle? 005 qfull 00200 3 3 0 tough math: k-representation of half exponential Extra keywords: (Mo-140:4.4) 2. At time zero, a wave function for a free particle in a zero-potential 1-dimensional space is: Ψ(x, 0) = Ae−|x|/ℓ eik0 x . a) Determine the normalization constant A. HINT: Remember it’s Ψ(x, 0)∗ Ψ(x, 0) that appears in the normalization equation. b) Sketch the x-space probability density |Ψ(x, 0)|2 . What is the e-folding distance of the probability density? NOTE: The e-folding distance is a newish term that means the distance in which an exponential function changes by a factor of e. It can be generalized to any function f (x) using the formula f (x) , xe = f (x)′

where xe is the generalized e-folding distance. The generalized e-folding distance is only locally valid to the region near the x where the functions are evaluated. The generalized e-folding distance is also sometimes called the scale height. If f (x) were an exponential function, xe would be the e-folding distance in the narrow sense. If f (x) were a linear function, xe would be the distance to the zero of the function.

c) Show that the wavenumber representation of free particle state is r 2ℓ 1 Ψ(k) = . π 1 + (k0 − k)2 ℓ2 This, of course, is the Fourier transform of Ψ(x, 0). Recall the wavenumber representation is time-independent since the wavenumber eigenstates are the stationary states of the potential. d) Confirm that Ψ(k) is normalized in wavenumber space. HINTS: You will probably need an integral table—unless you’re very, very good. Also remember it’s Ψ(k)∗ Ψ(k) that appears in the normalization integral; always easy to forget this when dealing with pure real functions. e) Write down the time-dependent solution Ψ(x, t) in Fourier transform form? Don’t try to evaluate the integral. What is ω in terms of k and energy E? 005 qfull 00300 3 3 0 tough math: Gaussian free wave packet spreading

38 Chapt. 5 Free Particles and Momemtum Representation Extra keywords: (Gr-50:2.22) 3. A free particle has an initial Gaussian wave function 2

Ψ(x, 0) = Ae−ax , where a and A are real positive constants. a) Normalize Ψ(x, 0). HINT: Recall that the integrand is |Ψ(x, t)|2 = Ψ(x, t)∗ Ψ(x, t). I’m always forgetting this myself when the function is pure real and there is no imaginary part to remind me of it. b) Determine the wavenumber representation Ψ(k) (which is time-independent). This involves a Gaussian integral where you have to complete the square in an exponential exponent. Note     2  b b b2 exp[−(ax2 + bx)] = exp −a x2 + x + 2 exp a 4a 4a2 "  2 #  2 b b = exp −a x + exp . 2a 4a
The exponetial factor exp b2 /4a comes out of the integral and the integral over the whole x-axis is just a simple Gaussian integral. c) Determine Ψ(x, t). You have to again do a Gaussian integral where you have to complete the square in an exponential exponent. It’s not that hard to do, but it is tedious and small errors can mess things up. d) Find the probability density |Ψ(x, t)|2 . This should be a Gaussian if all goes well. Sketch the function and identify the standard deviation σ. What happens to the probability density with time. HINT: Note the identities ∗  ∗    ac + i(bc − ad) + bd a + ib = exp exp c + id c2 + d2    ac − i(bc − ad) + bd = exp c2 + d2    a − ib = exp c − id and

√ ∗ √ ∗ √ ∗ √ a + ib = reiφ = reiφ/2 = re−iφ/2 ,

where a + ib magnitude and phase are r =

√

a2 + b2 and φ = tan−1 (b/a), respectively.

e) Find hxi, hx2 i, σx , hpi, hp2 i, and σp . HINT: These results follow immediately from the Gaussian nature of the functions in parts (d) and (b). f) Check that the Heisenberg uncertainty principle is satisfied. Does the equality ever hold? What’s true about the wave function at the time when the equality holds that is not true at other times? 005 qfull 00400 3 5 0 tough thinking: general free wave packet spreading Extra keywords: (CDL-342:4) 4. Consider a free particle in one dimension. a) Show using Ehrenfest’s theorem that hxi is linear in time and that hpi is constant.

Chapt. 5 Free Particles and Momemtum Representation 39 b) Write the equation of motion (time evolution equation) for hp2 i, then h[x, p]+ i (the subscript + indicates anticommutator), and then hx2 i: i.e., obtain expressions for the time derivatives of these quantities. Simplify the expressions for the derivatives as much as possible, but without loss of generality. You should get nice compact formal results. Integrate these derivatives with respect to time and remember constants of integration. c) Using the results obtained in parts (a) and (b) and for suitable choice of one of the constants of integration, show that

where ∆x2




0

and ∆p2




1 ∆x2 (t) = 2 ∆p2 0 t2 + ∆x2 0 m




are the initial standard deviations.

0

005 qfull 00500 2 5 0 moderate thinking: x-op and k-op in x and k representation 5. For a free particle in one dimension, the position and wavenumber representations of the state expanded in eigenfunctions of the other representation, are, respectively: Ψ(x, t) = and

∞

eikx φ(k, t) √ dk = 2π −∞

Z

Φ(k, t) = φ(k)e−iωt =

Z

∞

φ(k)

−∞

ei(kx−ωt) √ dk 2π

∞

 e−ikx Ψ(x, 0) √ dx e−iωt , 2π −∞

Z

where ω=

− h k2 . 2m

a) The expectation value for a general observable in Q the position representation is hQi =

Z

∞

Ψ(x, t)∗ QΨ(x, t) dx ,

−∞

where Q can depend on x and differentiations with respect to x, but not on k nor differentiations with respect to k. Substitute for Ψ(x, t) using the wavenumber expansion and find an expression that is a triple integral. b) The wavenumber observable in the position representation is 1 1 ∂ kop = − pop = , i ∂x h where pop is the momentum observable in the position representation. Using the result of the part (a) answer, identify the wavenumber observable in the wavenumber representation. c) The position observable xop in the position representation is just the coordinate x. Using the result of the part (a) answer, identify the position observable in the wavenumber representation. 005 qfull 00510 3 5 0 tough thinking: x-op and p-op in x and p representation Extra keywords: (Gr-117:3.51) a very general solution is given 6. In the position representation, the position operator xop is just x, a multiplicative variable. The momentum operator pop in the position representation is pdif =

− h ∂ . i ∂x

40 Chapt. 5 Free Particles and Momemtum Representation where we use the subsript “dif” here to indicate explicitly that this is a differentiating operator. a) Find the momentum operator pop in the momentum representation. HINTS: Operate with pdif on the Fourier transform expansion of a general wave function Ψ(x, t) =

−

∞

eipx/ h Ψ(p, t) √ dp −∞ 2π− h

Z

and work the components of the integral around (using whatever tricks you need) until you have Z ∞ − eipx/ h dp . pdif Ψ(x, t) = f (p, t) √ 2π− h −∞ The function f (p, t) is the Fourier transform of pdif Ψ(x, t) and operator acting on Ψ(p, t) to give you f (p, t) is the momentum operator in the momentum representation. b) Find the position operator xop in the momentum representation. HINTS: The same as for part (a), mutatis mutandis: find the Fourier transform of wave function xΨ(x, t), etc. c) What are the momentum representation versions of xk and pℓdif ? d) What is the momentum representation versions of xk pℓdif and pℓdif xk . By the by, you should remember how to interpret xk pℓdif and pℓdif xk : they are two successive operators acting on understood function to the right. Explicitly for a general function f (x), one could write

and

  xk pℓdif f (x) = xk pℓdif f (x)

  pℓdif xk f (x) = pℓdif xk f (x) ,

Unfortunately, there is sometimes ambiguity in writing formulae with operators: try to be clear. e) What is the momentum representation version of Q(x, pdif ) where Q is any linear combination of powers of x and pdif including mixed powers. HINTS: Consider the general term . . . xk pℓdif xm pndif and figure out what commutes with what. f) Show that the expectation value of Q is the same in both representations. Remember the Dirac delta function δ(k − k ′ ) =

Z

∞ −∞

′

ei(k−k )x dx . 2π

HINT:

Chapt. 6 Foray into Advanced Classical Mechanics

Multiple-Choice Problems 006 qmult 00100 1 1 2 easy memory: Newton’s 2nd law 1. Classical mechanics can be very briefly summarized by: a) b) c) d) e)

Newton’s Principia. Newton’s 2nd law. Lagrange’s Trait´e de m´ecanique analytique. Euler’s 80 volumes of mathematical works. Goldstein 3rd edition.

006 qmult 00200 1 4 1 easy deducto-memory: Lagrangian formulation Extra keywords: see GPS-12, 17, and 48 2. “Let’s play Jeopardy! For $100, the answer is: A formulation of classical mechanics that is usually restricted to systems with holonomic or semi-holonomic virtual-displacement workless constraints without dissipation and uses the function L = T − V .” a) b) c) d) e)

What What What What What

is is is is is

the the the the the

Lagrangian formulation, Alex? Hamiltonian formulation, Alex? Leibundgutian formulation, Alex? Harrisonian formulation, Alex? Sergeant Schultzian formulation, Alex?

006 qmult 00300 1 1 5 easy memory: Hamilton’s principle 3. A fruitful starting point for the derivation of Lagrange’s equations is: a) b) c) d) e)

Lagrange’s lemma. Newton’s scholium. Euler’s conjecture. Laplace’s hypothesis Hamilton’s principle.

Full-Answer Problems 006 qfull 01000 2 5 0 moderate thinking: Lorentz force 1. The Lorentz force   ~v ~ ~ ~ F =q E+ ×B c (here expressed in Gaussian units: Ja-238) can be obtained from Lagrange’s equations using a Lagrangian containing a generalized potential   ~v ~ U =q φ− ·A , c 41

42 Chapt. 6 Foray into Advanced Classical Mechanics ~ is the vector potential of electromagnetism. The where φ is the electric potential and A Lagrangian is L = T − U , where T is the kinetic energy. Lagrange’s equations are d dt



∂L ∂ q˙i



−

∂L =0, ∂qi

where qi is a generalized coordinate (not charge) and q˙i is the total time derivative of q: (i.e., the rate of change of qi which describes an actual particle. Work from the Lorentz force expression for component i to ∂U d Fi = − + ∂xi dt



∂U ∂ x˙ i



,

where the xi are the Cartesian coordinates of a particle (x˙ i are the particle velocity components). Then verify that m¨xi = Fi follows from the Lagrange equations. You may need some hints. Recall that ~ ~ = −∇φ − 1 ∂ A E c ∂t

and

~ B =∇×A

(Ja-219). The Levi-Civita symbol εijk will be useful since (∇ × A)i = εijk

∂ Ak . ∂xj

where Einstein summation has been used. Recall ( 1, ijk cyclic; εijk = −1, ijk anticyclic; 0, if two indices the same. The identity (with Einstein summation) εijk εiℓm = δjℓ δkm − δjm δkℓ is also useful. I’ve never found an elegant derivation of this last identity: the only proof seems to be by exhaustion. Note also that the total time derivative is interpreted as the rate of change of a quantity as the particle moves. Thus dAi ∂Ai ∂Ai ∂Ai dxj ∂Ai ∂Ai ∂Ai x˙ j = vj , = + = + + dt ∂t ∂xj dt ∂t ∂xj ∂t ∂xj where we again use Einstein summation.

Chapt. 7 Linear Algebra

Multiple-Choice Problems 007 qmult 00100 1 1 5 easy memory: vector addition 1. The sum of two vectors belonging to a vector space is: a) b) c) d) e)

a scalar. another vector, but in a different vector space. a generalized cosine. the Schwarz inequality. another vector in the same vector space.

007 qmult 00200 1 4 4 easy deducto-memory: Schwarz inequality 2. “Let’s play Jeopardy! For $100, the answer is: |hα|βi|2 ≤ hα|αihβ|βi.” What is

, Alex?

a) the triangle inequality b) the Heisenberg uncertainty principle c) Fermat’s last theorem d) the Schwarz inequality e) Schubert’s unfinished last symphony 007 qmult 00300 1 4 5 easy deducto-memory: Gram-Schmidt procedure 3. Any set of linearly independent vectors can be orthonormalized by the: a) Pound-Smith procedure. b) Li Po tao. c) Sobolev method. d) Sobolev-P method. e) Gram-Schmidt procedure. 007 qmult 00400 1 4 4 moderate memory: definition unitary matrix 4. A unitary matrix is defined by the expression: a) U = U T , where superscript T means transpose. d) U −1 = U † . e) U −1 = U ∗ .

b) U = U † .

c) U = U ∗ .

007 qmult 00500 2 3 4 moderate math: trivial eigenvalue problem 5. What are the eigenvalues of   1 −i ? i 1 a) Both are 0.

b) 0 and 1.

c) 0 and −1.

d) 0 and 2.

007 qmult 00600 1 4 5 moderate memory: riddle Hermitian matrix 6. Holy peccant poets Batman, it’s the Riddler. I charge to the right and hit on a ket, and if it’s not eigen, it’s still in the set, I charge to left and with a quick draw make a new bra from out of a bra. Not fish nor fowl nor quadratic, not uncanny tho oft Q-mechanic, 43

e) −i and 1.

44 Chapt. 7 Linear Algebra and transposed I’m just the right me if also complexicated as you can see. My arrows down drawn from quivered, the same when sped to the world delivered aside from a steady factor, rock of reality, mayhap of a quantum and that’s energy. a) b) c) d) e)

A unitary operator. A ket—no, no, a bra vector. An eigenvalue. Hamlet. A Hermitian matrix.

Full-Answer Problems 007 qfull 00090 1 5 0 easy thinking: ordinary vector space Extra keywords: (Gr-77:3.1) 1. Consider ordinary 3-dimensional vectors with complex components specified by a 3-tuple: (x, y, z). They constitute a 3-dimensional vector space. Are the following subsets of this vector space vector spaces? If so, what is their dimension? HINT: See Gr-435 for all the properties a vector space must have. a) The subset of all vectors (x, y, 0). b) The subset of all vectors (x, y, 1). c) The subset of all vectors of the form (a, a, a), where a is any complex number. 007 qfull 00100 2 5 0 moderate thinking: vector space, polynomial Extra keywords: (Gr-78:3.2) 2. A vector space is constituted by a set of vectors {|αi, |βi, |γi, . . .} and a set of scalars {a, b, c, . . .} (ordinary complex numbers is all that quantum mechanics requires) subject to two operations vector addition and scalar multiplication obeying the certain rules. Note it is the relations between vectors that make them constitute a vector space. What they “are” we leave general. The rules are: i) A sum of vectors is a vector: |αi + |βi = |γi , where |αi and |βi are any vectors in the space and |γi also in the space. Note we have not defined what vector addition consists of. That definition goes beyond the general requirements. ii) Vector addition is commutative: |αi + |βi = |βi + |αi . iii) Vector addition is associative: (|αi + |βi) + |γi = |αi + (|βi + |γi) . iv) There is a zero or null vector |0i such that |αi + |0i = |αi ,

Chapt. 7 Linear Algebra 45 v) For every vector |αi there is an inverse vector | − αi such that |αi + | − αi = |0i . vi) Scalar multiplication of a vector gives a vector: a|αi = |βi . vii) Scalar multiplication is distributive on vector addition: a(|αi + |βi) = a|αi + a(|βi) . viii) Scalar multiplication is distributive on scalar addition: (a + b)|αi = a|αi + b|αi . ix) Scalar multiplication is associative with respect to scalar multiplication: (ab)|αi = a(b|αi) . x) One has 0|αi = |0i . xi) Finally, one has 1|αi = |αi . NOTE: Note that |0i = 0|αi = [1 + (−1)]|αi = |αi + (−1)|αi , and thus we find that | − αi = −|αi . So the subtraction of a vector is just the addition of its inverse. This is consistent with all ordinary math. If any vector in the space can be written as linear combination of a set of linearly independent vectors, that set is called a basis and is said to span the set. The number of vectors in the basis is the dimension of the space. In general there will be infinitely many bases for a space. Finally the question. Consider the set of polynomials {P (x)} (with complex coefficients) and degree less than n. For each of the subsets of this set (specified below) answer the following questions: 1) Is the subset a vector space? Inspection usually suffices to answer this question. 2) If not, what property does it lack? 3) If yes, what is the most obvious basis and what is the dimension of the space? a) The subset that is the whole set. b) The subset of even polynomials. c) The subset where the highest term has coefficient a (i.e., the leading coefficient is a) and a is a general complex number, except a 6= 0. d) The subset where P (x = g) = 0 where g is a general real number. (To be really clear, I mean the subset of polynomials that are equal to zero at the point x = g.)

46 Chapt. 7 Linear Algebra e) The subset where P (x = g) = h where g is a general real number and h is a general complex number, except h 6= 0. 007 qfull 00110 2 5 0 moderate thinking: unique expansion in basis Extra keywords: (Gr-78:3.3) 3. Prove that the expansion of a vector in terms of some basis is unique: i.e., the set of expansion coefficients for the vector is unique. 007 qfull 00200 3 5 0 tough thinking: Gram-Schmidt orthonormalization Extra keywords: (Gr-79:3.4) 4. Say {|αi i} is a basis (i.e., a set of linearly independent vectors that span a vector space), but it is not orthonormal. The first step of the Gram-Schmidt orthogonalization procedure is to normalize the (nominally) first vector to create a new first vector for a new orthonormal basis: |α′1 i =

|α1 i , ||α1 ||

where recall that the norm of a vector |αi is given by p ||α|| = || |α1 i || = hα|αi .

The second step is create a new second vector that is orthogonal to the new first vector using the old second vector and the new first vector: |α′2 i =

|α2 i − |α′1 ihα′1 |α2 i . || |α2 i − |α′1 ihα′1 |α2 i ||

Note we have subtracted the projection of |α2 i on |α′1 i from |α2 i and normalized. a) Write down the general step of the Gram-Schmidt procedure. b) Why must an orthonormal set of non-null vectors be a linearly independent. c) Is the result of a Gram-Schmidt procedure independent of the order the original vectors are used? HINT: Say you first use vector |αa i of the old set in the procedure. The first new vector is just |αa i normalized: i.e., |α′a i=|αa i/||αa ||. All the other new vectors will be orthogonal to |α′a i. But what if you started with |αb i which in general is not orthogonal to |αa i? d) How many orthonormalized bases can an n dimensional space have in general? (Ignore the strange n = 1 case.) HINT: Can’t the Gram-Schmidt procedure be started with any vector at all in the vector space? e) What happens in the procedure if the original vector set {|αi i} does not, in fact, consist of all linearly independent vectors? To understand this case analyze another apparently different case. In this other case you start the Gram-Schmidt procedure with n original vectors. Along the way the procedure yields null vectors for the new basis. Nothing can be done with the null vectors: they can’t be part of a basis or normalized. So you just put those null vectors and the vectors they were meant to replace aside and continue with the procedure. Say you got m null vectors in the procedure and so ended up with n − m non-null orthonormalized vectors. Are these n − m new vectors independent? How many of the old vectors were used in constructing the new n − m non-null vectors and which old vectors were they? Can all the old vectors be recontructed from the the new n− m non-null vectors? Now answer the original question. f) If the original set did consist of n linearly independent vectors, why must the new orthonormal set consist of n linearly independent vectors? HINT: Should be just a corollary of the part (e) answer.

Chapt. 7 Linear Algebra 47 g) Orthonormalize the 3-space basis consisting of     1+i i |α1 i =  1  , |α2 i =  3  , i 1

and



 0 |α3 i =  32  . 0

Input the vectors into the procedure in the reverse of their nominal order: why might a marker insist on this? Note setting kets equal to columns is a lousy notation, but you-all know what I mean. The bras, of course, should be “equated” to the row vectors. HINT: Make sure you use the normalized new vectors in the construction procedure.

007 qfull 00300 2 3 0 moderate math: prove the Schwarz inequality Extra keywords: (Gr-80:3.5) 5. As Andy Rooney says (or used to say if this problem has reached the stage where only old fogies remember that king of the old fogies) don’t you just hate magic proofs where you start from some unmotivated expression and do a number of unmotivated steps to arrive at a result that you could never have been guessed from the way you were going about getting it. Well sans too many absurd steps, let us see if we can prove the Schwarz inequality |hα|βi|2 ≤ hα|αihβ|βi for general vectors |αi and |βi. Note the equality only holds in two cases. First when |βi = a|αi, where a is some complex constant. Second, when either or both of |αi and |βi are null vectors: in this case, one has zero equals zero. NOTE: A few facts to remember about general vectors and inner products. Say |αi and |βi are general vectors. By the definition of the inner product, we have that hα|βi = hβ|αi∗ . This implies that hα|αi is pure real. If c is a general complex number, then the inner product of |αi and c|βi is hα|c|βi = chα|βi. Next we note that that another inner-product property is that p hα|αi ≥ 0 and the equality only holds if |αi is the null vector. The norm of |αi is ||α|| = hα|αi and |αi can be normalized if it is not null: i.e., for |αi not null, the normalized version is |ˆ αi = |αi/||α||. a) In doing the proof of the Schwarz inequality, it is convenient to have the result that the bra c∗ hγ| = hγ|c∗ corresponds the ket c|γi, where |γi is a general vector and c is a general complex number. Prove this correspondance. HINT: Define |ǫi = c|γi, take the inner product with general vector |δi, and do some manipulation making use of general vector and inner-product properties. b) The next thing to do is to figure out what the Schwarz inequality is saying about vectors including those 3-dimensional things we have always called vectors. Let us a restrict the generality of |αi by demanding it not be a null vector for which the Schwarz inequality is already proven. Since |αi is not null, it can be normalized. Let |ˆ αi = |αi/||α|| be the normalized version of |αi. Divide the Schwarz inequality by ||α||2 . Now note that the component of |βi along the |ˆ αi direction is |βk i = |ˆ αihˆ α|βi . Evaluate hβk |βk i. Now what is the Schwarz inequality telling us. c) The vector component of |βi that is orthogonal to |ˆ αi (and therefore |βk i) is |β⊥ i = |βi − |βk i . Prove this and then prove the Schwarz inquality itself (for |αi not null) by evaluating hβ|βi expanded in components. Schwarz inequality for |αi not a null vector. What if |αi is a null vector?

48 Chapt. 7 Linear Algebra 007 qfull 00310 1 3 0 easy math: find a generalized angle Extra keywords: (Gr-80:3.6) 6. The general inner-product vector space definition of generalized angle according to Gr-440 is |hα|βi| cos θgen = p , hα|αihβ|βi

where |αi and |βi are general non-zero vectors.

a) Is this definition completely consistent with the ordinary definition of an angle from the ordinary vector dot product? Why or why not? b) Find the generalized angle between vectors 

 1+i |αi =  1  i

and



 4−i |βi =  0  . 2 − 2i

007 qfull 00400 1 3 0 easy math: prove triangle inequality Extra keywords: (Gr-80:3.7) 7. Prove the triangle inequality: ||(|αi + |βi)|| ≤ ||α|| + ||β|| . HINT: Start with ||(|αi + |βi)||2 , expand, and use reality and the Schwarz inequality |hα|βi|2 ≤ hα|αihβ|βi = ||α||2 × ||β||2 . 007 qfull 00500 3 3 0 tough math: simple matrix identities Extra keywords: (Gr-87:3.12) 8. Prove the following matrix identities: a) (AB)T = B T AT , where superscript “T” means transpose. b) (AB)† = B † A† , where superscript † means Hermitian conjugate. c) (AB)−1 = B −1 A−1 .

d) (U V )−1 = (U V )† (i.e., U V is unitary) given that U and V are unitary. In other words, prove the product of unitary matrices is unitiary. e) (AB)† = AB (i.e., AB is Hermitian) given that A and B are commuting Hermitian matrices. Does the converse hold: i.e., does (AB)† = AB imply A and B are commuting Hermitian matrices? HINTS: Find a trivial counterexample. Try B = A−1 . f) (A + B)† = A + B (i.e., A + B is Hermitian) given that A and B are Hermitian. Does the converse hold? HINT: Find a trivial counterexample to the converse. g) (U + V )† = (U + V )−1 (i.e., U + V is unitary) given that U and V are unitary—that is, prove this relation if it’s indeed true—if it’s not true, prove that it’s not true. HINT: Find a simple counterexample: e.g., two 2 × 2 unit matrices. 007 qfull 00510 2 5 0 moderate thinking: commuting operations Extra keywords: (Gr-84)

Chapt. 7 Linear Algebra 49 9. There are 4 simple operations that can be done to a matrix: inversing, (−1), complex conjugating (∗), transposing (T ), and Hermitian conjugating (†). Prove that all these operations mutually commute. Do this systematically: there are   4! 4 = =6 2 2!(4 − 2)! combinations of the 2 operations. We assume the matrices have inverses for the proofs involving them. 007 qfull 00600 3 3 0 tough math: basis change results Extra keywords: (Gr-87:3.14) 10. Do the following. a) Prove that matrix multiplication is preserved under similarity or linear basis change: i.e., if Ae Be = Ce in the e-basis, then Af Bf = Cf in the f -basis where S is the basis change matrix from e-basis to the f -basis. Basis change does not in general preserve symmetry, reality, or Hermiticity. But since I don’t want to find the counterexamples, I won’t ask you to. b) If He in the e-basis is a Hermitian matrix and the basis change to the f -basis U is unitary, prove that Hf is Hermitian: i.e., Hermiticity is is preserved. c) Prove that basis orthonormality is preserved through a basis change U iff (if and only if) U is unitary. 007 qfull 00700 2 5 0 moderate thinking: square-integrable, inner product Extra keywords: no analog Griffiths’s problem, but discussion Gr-95–6, Gr2005-94–95 11. If f (x) and g(x) are square-integrable complex functions, then the inner product Z ∞ hf |gi = f ∗ g dx −∞

exists: i.e., is convergent to a finite value. In other words, that f (x) are g(x) are squareintegrable is sufficient for the inner product’s existence. a) Prove the statement for the case where f (x) and g(x) are real functions. HINT: In doing this it helps to define a function  f (x) where |f (x)| ≥ |g(x)| (which we call the f region); h(x) = g(x) where |f (x)| < |g(x)| (which we call the g region), and show that it must be square-integrable. Then “squeeze” hf |gi. b) Now prove the statement for complex f (x) and g(x). HINTS: Rewrite the functions in terms of their real and imaginary parts: i.e., f (x) = fRe (x) + ifIm (x) and g(x) = gRe (x) + igIm (x) . Now expand hf |gi =

Z

∞

f ∗ g dx

−∞

in the terms of the new real and imaginary parts and reduce the problem to the part (a) problem.

50 Chapt. 7 Linear Algebra c) Now for the easy part. Prove the converse of the statement is false. HINT: Find some trivial counterexample. d) Now another easy part. Say you have a vector space of functions {fi } with inner product defined by Z ∞ fj∗ fk dx . −∞

Prove the following two statements are equivalent: 1) the inner product property holds; 2) the functions are square-integrable. 007 qfull 00800 2 3 0 moderate math: Gram-Schmidt, Legendre polynomials Extra keywords: (Gr-96:3.25) 12. The Gram-Schmidt procedure can be used to construct a set of orthonormal vectors by linear combination from a set of linearly independent, but non-orthonormal, vectors. It is a sort of brute force approach to use when more elegant methods of orthonormalization are not available. The Gram-Schmidt procedure is as follows. Say {|φn i} are a set of N linearly independent vectors that are NOT assumed to be orthonormal. One has ordered them in some reasonable manner indicated by the index n which increases from 1 to N . From this set we construct the orthonormal set {|ˆ un i} of N normalized vectors where the hat symbol indicates their normalization. The unnormalized nth vector of the new set is given by |un i = |φn i −

n−1 X ℓ=1

|ˆ uℓ ihˆ uℓ |φn i ,

where the sum vanishes for n = 1, and the corresponding normalized vector is given by |ˆ un i = where ||un || =

|un i , ||un ||

p hun |un i is the norm of |un i.

a) Prove the Gram-Schmidt procedure by induction. b) In general there is an uncountable infinity of orthonormal sets that can be constructed from a non-orthonormal set. For example, consider ordinary vectors in 3-dimensional Euclidean space. By rotation of a orthonormal set of unit vectors an uncountable infinity of orthonormal sets of unit vectors can be created. Different orthonormal sets can be created using the Gram-Schmidt procedure just by itself. Say that there are N vectors in linearly independent non-orthonomcal set and each vector is orthogonal with all the others in the set. Show that you can create at least N different orthogonal sets by the Gram-Schmidt procedure. c) Using the Gram-Schmidt procedure on the interval [−1, 1] (with weight function w = 1 in the integral rule for the inner product) find the first three orthonormalized polynomial vectors starting from the linearly independent, but non-orthonomal set of polynomial vectors given by φn = xn for integer n ∈ [0, ∞). Show that the orthonormalized set are the first three normalized Legendre polynomials {Pˆn (x)}. The normalized Legendre polynomials are given by r 2n + 1 Pn (x) , Pˆn (x) = 2 where Pn (x) is a standard Legendre polynomial (e.g., Ar-547; WA-569). (Note an arbitrary phase factor eiθ can also be multiplied to a normalized vector.) The reason why the Legendre polynomials arn’t normalized is that the standard forms are what one gets straight from the generating function. The generating function approach to the Legendre polynomials allows you to prove many of their properties quickly (e.g., Ar-534, WA-553).

Chapt. 7 Linear Algebra 51 Table: Legendre Polynomials Order n

Pn

0

P0 = 1

1

P1 = x

2

P2 = (1/2)(3x2 − 1)

3 4 5

P3 = (1/2)(5x3 − 3x)

P4 = (1/8)(35x4 − 30x2 + 3)

P5 = (1/8)(63x5 − 70x3 + 15x)

Note—The polynomials are from Ar-541 and WA-554. The degree of a Legendre polynomial is given by its order number n (WA-557). d) The normalized Legendre polynomials form a complete orthonormal basis for piecewise continuous, normalizable functions in the interval [−1, 1] (Ar-443). For a general polynomial of degree n, show that the expansion Qn (x) =

∞ X ℓ=0

Pˆℓ (x)hPˆℓ |Qn i

actually truncates at ℓ = n. HINT: Consider xn and how could construct it from Legendre polynomials starting with Pn (x). e) The part (c) result suggests that if the Gram-Schmidt procedure is continued beyond the first 3 vectors of the set {φn }, one will continue getting the normalized Legendre polynomials in order. Prove this is so by comparing the Gram-Schmidt procedure result for hPˆk |un i and the expansion for |un i in the set {|Pˆn i}. In evaluating hPˆk |un i, assume that the |ˆ uℓ i for ℓ < n are |Pˆℓ i. HINT: The part (d) result is also needed for the proof. 007 qfull 00900 1 3 0 easy math: verifying a sinusoidal basis Extra keywords: (Gr-96:3.26) 13. Consider the set of trigonometric functions defined by f (x) =

N X

[an sin(nx) + bn cos(nx)]

n=0

on the interval [−π, π]. Show that the functions defined by 1 φk (x) = √ eikx , 2π

where

k = 0, ±1, ±2, . . . , ±N

are an orthonormal basis for the trigonometric set. What is the dimension of the space spanned by the basis? 007 qfull 01000 2 3 0 moderate math: reduced SHO operator, Hermiticity Extra keywords: (Gr-99:3.28), dimensionless simple harmonic oscillator Hamiltonian 14. Consider the operator d2 Q = − 2 + x2 . dx a) Show that f (x) = e−x

2

/2

is an eigenfunction of Q and determine its eigenvalue.

52 Chapt. 7 Linear Algebra b) Under what conditions, if any, is Q a Hermitian operator? HINTS: Recall hg|Q† |f i∗ = hf |Q|gi is the defining relation for the Hermitian conjugate Q† of operator Q. You will have to write the matrix element hf |Q|gi in the position representation and use integration by parts to find the conditions. 007 qfull 01100 2 5 0 moderate thinking: Hilbert space problems Extra keywords: (Gr-103:3.33) 15. Do the following. a) Show explicitly that any linear combination of two functions in the Hilbert space L2 (a, b) is also in L2 (a, b). (By explicitly, I mean don’t just refer to the definition of a vector space which, of course requires the sum of any two vectors to be a vector.) b) For what values of real number s is f (x) = |x|s in L2 (−a, a)

c) Show that f (x) = e−|x| is in L2 = L2 (−∞, ∞). Find the wavenumber space representation of f (x): recall the wavenumber “orthonormal” basis states in the position representation are eikx hx|ki = √ . 2π

007 qfull 01200 2 5 0 moderate thinking: Hermitian conjugate of AB 16. Some general operator and vector identities should be proven. Recall the definition of the Hermitian conjugate of general operator Q is giveny by hα|Q|βi = hβ|Q† |αi∗ , where |αi and |βi are general vectors.

a) Prove that the bra corresponding to vector Q|βi is hβ|Q† for general Q and |βi. HINT: Define |β ′ i = Q|βi and then take the inner product of that vector with another general vector |αi and use the definition of the Hermitian conjugate. I’d use bra hβ ′ | on ket |αi and then reverse the order complex conjugating.

b) Show that the Hermitian conjugate of a scalar c is just its complex conjugate. c) Prove for operators, not matrices, that (AB)† = B † A† . The result is, of course, consistent with matrix representations of these operators. But there are representations in which the operators are not matrices: e.g., the momentum operator in the position representation is differentiating operator p=

− h ∂ . i ∂x

Our proof holds for such operators too since we’ve done the proof in the general operatorvector formalism. d) Generalize the proof in part (c) for an operator product of any number. e) Prove that (A + B)† = A† + B † . f) Prove that c[A, B] is a Hermitian operator for Hermitian A and B only when c is pure imaginary constant.

Chapt. 7 Linear Algebra 53 007 qfull 01300 3 5 0 tough thinking: operators and matrices isomorphism 17. Expressions involving vector linear transformations or operators often (always?) isomorphic to the corresponding matrix expressions when the operators are represented by matrices in particular orthonormal bases. We would like to demonstrate this statement for a few important ˆ and leave the corresponding simple cases. For clarity express operators with hats (e.g., A) matrices unadorned (e.g., A). Consider a general orthonormal basis {|ii} where i serves as a labeling index. Recall that the unit operator using this basis is I = |iihi| , where we use the Einstein summation rule, and so there is a sum on i. Recall that the ijth matrix element of A is defined by ˆ . Aij = hi|A|ji ˆ This definition means that the scalar product ha|A|bi, where |ai and |bi are general vectors, can be reexpressed by matrix expression: ˆ = ha|iihi|A|jihj|bi ˆ ha|A|bi = a∗i Aij bj = ~a† A~b , where ~a and ~b are column vector n-tuples and where we have used the Einstein rule. Prove that the following operator expressions are isomorphic to their corresponding matrix expressions. ˆ a) Sum of operators Aˆ + B. ˆ b) Product of operators AˆB. c) Hermitian conjugation Aˆ† . ˆ †=B ˆ † Aˆ† . d) The identity (AˆB) 007 qfull 01400 2 5 0 moderate thinking: bra ket projector completeness Extra keywords: (Gr-118:3.57) See also CDL-115, 138 18. For an inner product vector space there is some rule for calculating the inner product of two general vectors: an inner product being a complex scalar. If |αi and |βi are general vectors, then their inner product is denoted by hα|βi , where in general the order is significant. Obviously different rules can be imagined for a vector space which would lead to different values for the inner products. But the rule must have three basic properties: (1) (2) and (3)

hβ|αi = hα|βi∗ , hα|αi ≥ 0 , where hα|αi = 0 if and only if |αi = |0i,   hα| b|βi + c|γi = bhα|βi + chα|γi ,

where |αi, |βi, and |γi are general vectors of the vector space and b and c are general complex scalars. There are some immediate corollaries of the properties. First, if hα|βi is pure real, then hβ|αi = hα|βi .

54 Chapt. 7 Linear Algebra Second, if hα|βi is pure imaginary, then hβ|αi = −hα|βi . Third, if |δi = b|βi + c|γi , then hδ|αi∗ = hα|δi = bhα|βi + chα|γi which implies hδ|αi = b∗ hβ|αi + c∗ hγ|αi . This last result makes



 hβ|b + hγ|c |αi = b∗ hβ|αi + c∗ hγ|αi ∗

∗

a meaningful expression. The 3rd rule for a vector product inner space and last corollary together mean that the distribution of inner product multiplication over addition happens in the normal way one is used to. Dirac had the happy idea of defining dual space vectors with the notation hα| for the dual vector of |αi: hα| being called the bra vector or bra corresponding to |αi, the ket vector or ket: “bra” and “ket” coming from “bracket.” Mathematically, the bra hα| is a linear function of the vectors. It has the property of acting on a general vector |βi and yielding a complex scalar: the scalar being exactly the inner product hα|βi. One immediate consequence of the bra definition can be drawn. Let |αi, |βi, and a be general and let |α′ i = a|αi . Then hα′ |βi = hβ|α′ i∗ = a∗ hβ|αi∗ = a∗ hα|βi implies that the bra corresponding to |α′ i is given by hα′ | = a∗ hα| = hα|a∗ . The use of bra vectors is perhaps unnecessary, but they do allow some operations and properties of inner product vector spaces to be written compactly and intelligibly. Let’s consider a few nice uses. a) The projection operator or projector on to unit vector |ei is defined by Pop = |eihe| . This operator has the property of changing a vector into a new vector that is |ei times a scalar. It is perfectly reasonable to call this new vector the component of the original vector in the direction of |ei: this definition of component agrees with our 3-dimensional Euclidean definition of a vector component, and so is a sensible generalization of that the 3-dimensional Euclidean definition. This generalized component would also be the contribution of a basis of which |ei is a member to the expansion of the original vector: again the usage of the word component is entirely reasonable. In symbols Pop |αi = |eihe|αi = a|ei , where a = he|αi.

Chapt. 7 Linear Algebra 55 2 n Show that Pop = Pop , and then that Pop = Pop , where n is any integer greater than or equal to 1. HINTS: Write out the operators explicitly and remember |ei is a unit vector.

b) Say we have Pop |αi = a|αi , where Pop = |eihe| is the projection operator on unit vector |ei and |αi is unknown non-null vector. Solve for the TWO solutions for a. Then solve for the |αi vectors corresponding to these solutions. HINTS: Act on both sides of the equation with he| to find an equation for one a value. This equation won’t yield the 2nd a value—and that’s the hint for finding the 2nd a value. Substitute the a values back into the original equation to determine the corresponding |αi vectors. Note one a value has a vast degeneracy in general: i.e., many vectors satisfy the original equation with that a value. c) The Hermitian conjugate of an operator Q is written Q† . The definition of Q† is given by the expression hβ|Q† |αi = hα|Q|βi∗ , where |αi and |βi are general vectors. Prove that the bra corresponding to ket Q|βi must hβ|Q† for general |αi. HINTS: Let |β ′ i = Q|βi and substitute this for Q|βi in the defining equation of the Hermitian conjugate operator. Note operators are not matrices (although they can be represented as matrices in particular bases), and so you are not free to use purely matrix concepts: in particular the concepts of tranpose and complex conjugation of operators are not generally meaningful. d) Say we define a particular operator Q by Q = |φihψ| , where |φi and |ψi are general vectors. Solve for Q† . Under what condition is Q† = Q ? When an operator equals its Hermitian conjugate, the operator is called Hermitian just as in the case of matrices. e) Say {|ei i} is an orthonormal basis. Show that |ei ihei | = 1 , where we have used Einstein summation and 1 is the unit operator. HINT: Expand a general vector |αi in the basis.

Chapt. 8 Operators, Hermitian Operators, Braket Formalism

Multiple-Choice Problems 008 qmult 00090 1 4 5 easy deducto-memory: example Hilbert space 1. “Let’s play Jeopardy! For $100, the answer is: A space of all square-integrable functions on the x interval (a, b).” What is a

, Alex?

a) non-inner product vector space b) non-vector space d) Dogbert space e) Hilbert space

c) Dilbert space

008 qmult 00100 1 1 3 easy memory: complex conjugate of scalar product 2. The scalar product hf |gi∗ in general equals: a) hf |gi.

b) ihf |gi.

c) hg|f i.

d) hf |i|gi.

e) hf |(−i)|gi.

008 qmult 00200 1 4 3 easy deducto-memory: what operators do 3. “Let’s play Jeopardy! For $100, the answer is: It changes a vector into another vector.” What is a/an

, Alex?

a) wave function b) scalar product e) telephone operator

c) operator

d) bra

008 qmult 00300 2 1 5 moderate memory: Hermitian conjugate of product 4. Given general operators A and B, (AB)† equals: a) AB. b) A† B † .

c) A.

e) B † A† .

d) B.

008 qmult 00400 2 5 5 moderate thinking: general Hermitian conjugation 5. The Hermitian conjugate of the operator λ|φihχ|ψihℓ|A (with λ a scalar and A an operator) is: a) λ|φihχ|ψihℓ|A. e) A† |ℓihψ|χihφ|λ∗ .

b) λ|φihχ|ψihℓ|A† .

c) A|ℓihψ|χihφ|λ∗ .

d) A|ℓihψ|χihφ|λ.

008 qmult 00500 1 1 5 easy memory: compatible observables 6. Compatible observables: a) anticommute. b) are warm and cuddly with each other. c) have no hair. d) have no complete simultaneous orthonormal basis. e) commute. 008 qmult 00600 1 1 3 easy memory: parity operator 7. The parity operator Π acting on f (x) gives: df /dx.

b) 1/f (x).

c) f (−x).

d) 0.

e) a spherical harmonic.

008 qmult 00700 1 4 3 easy deducto-memory: braket expectation value 8. Given the position representation for an expectation value hQi =

Z

∞

Ψ(x)∗ QΨ(x) dx ,

−∞

56

Chapt. 8 Operators, Hermitian Operators, Braket Formalism 57 what is the braket representation? a) hQ|Ψ∗ |Qi.

b) hΨ∗ |Q|Ψi.

c) hΨ|Q|Ψi.

d) hΨ|Q† |Ψi.

e) hQ|Ψ|Qi.

008 qmult 00800 1 4 3 easy deducto-memory: Hermitian eigenproblem 9. What are the three main properties of the solutions to a Hermitian operator eigenproblem? a) (i) The eigenvalues are pure IMAGINARY. (ii) The eigenvectors are guaranteed orthogonal, except for those governed by degenerate eigenvalues and these can always be orthogonalized. (iii) The eigenvectors DO NOT span all space. b) (i) The eigenvalues are pure REAL. (ii) The eigenvectors are guaranteed orthogonal, except for those governed by degenerate eigenvalues and these can always be orthogonalized. (iii) The eigenvectors span all space in ALL cases. c) (i) The eigenvalues are pure REAL. (ii) The eigenvectors are guaranteed orthogonal, except for those governed by degenerate eigenvalues and these can always be orthogonalized. (iii) The eigenvectors span all space for ALL FINITE-DIMENSIONAL spaces. In infinite dimensional cases they may or may not span all space. It is quantum mechanics postulate that the eigenvectors of an observable (which is a Hermitian operator) span all space. d) (i) The eigenvalues are pure IMAGINARY. (ii) The eigenvectors are guaranteed orthogonal, except for those governed by degenerate eigenvalues and these can always be orthogonalized. (iii) The eigenvectors span all space in ALL FINITE-DIMENSIONAL spaces. In infinite dimensional cases they may or may not span all space. e) (i) The eigenvalues are pure REAL. (ii) The eigenvectors are guaranteed orthogonal, except for those governed by degenerate eigenvalues and these can always be orthogonalized. 008 qmult 00900 1 4 5 easy deducto-memory: definition observable 10. “Let’s play Jeopardy! For $100, the answer is: A physically significant Hermitian operator possessing a complete set of eigenvectors.” What is a/an a) conjugate

, Alex? b) bra

c) ket

d) inobservable

e) observable

008 qmult 01000 1 4 4 easy deducto-memory: time-energy inequality 11. In the precisely-formulated time-energy inequality, the ∆t is: a) b) c) d) e)

the standard deviation of time. the standard deviation of energy. a Hermitian operator. the characteristic time for an observable’s value to change by one standard deviation. the characteristic time for the system to do nothing.

008 qmult 02000 1 1 5 easy memory: common eigensets Extra keywords: See CDL-140ff 12. The statements “two observables commute” and “a common eigenset can be constructed for two observables” are in flat contradiction. b) unrelated. d) irrelevant in relation to each other. the other.

Full-Answer Problems 008 qfull 00010 1 1 0 easy memory: what is a ket?

c) in non-intersecting Venn diagrams. e) are equivalent in the sense that one implies

58 Chapt. 8 Operators, Hermitian Operators, Braket Formalism 1. What is a ket (representative general symbol |Ψi)? 008 qfull 00015 1 1 0 easy memory: what is a bra? 2. What is a bra? (Representative general symbol hΨ|.) 008 qfull 00020 1 1 0 easy memory: why the braket formalism? 3. Why is quantum mechanics at the advanced level formulated in the braket formalism? 008 qfull 00030 2 5 0 moderate thinking: Hermiticity and expectation values Extra keywords: (Gr-94:3.21) 4. Recall the definition of Hermitian conjugate for a general operator Q is hα|Q|βi = hβ|Q† |αi∗ , where |αi and |βi are general vectors. If Q is Hermitian, Q† = Q : i.e., Q is its own Hermitian conjugate. a) If Q is Hermitian, prove that the expectation value of a general vector |γi, hγ|Q|γi , is pure real. b) If the expectation value hγ|Q|γi is always pure real for general |γi, prove that Q is Hermitian. The statement to be proven is the converse of the statement in part (a). HINT: First show that hγ|Q|γi = hγ|Q† |γi . Then let |αi and |βi be general vectors and construct a vector |ξi = |αi + c|βi, where c is a general complex scalar. Note that the bra corresponding to c|βi is c∗ hβ|. Expand both sides of hξ|Q|ξi = hξ|Q† |ξi , and then keep simplifying both sides making use of the of the first thing proven and the definition of a Hermitian conjugate. Note that the It may be useful to note that A† )† = A

and

(A + B)† = A† + B † ,

where A and B are general operators and You should be able to construct an expression where choosing c = 1 and then c = i requires Q = Q† . c) What simple statement follows from the proofs in parts (a) and (b)? 008 qfull 00032 2 5 0 moderate thinking: energy operator 5. We usually think of the Hamiltonian as being the “energy” operator, but term energy operator is also used for the operator ∂ Eop = i− h ∂t (Mo-184). a) Find general expressions for the eigenvalues and normalizable eigenstates of Eop .

Chapt. 8 Operators, Hermitian Operators, Braket Formalism 59 b) Show that Eop is not, in fact, a Hermitian operator in the space of physical wave functions. HINT: Recall the Hermitian conjugate of an operator Q is defined by hα|Q† |βi = hβ|Q|αi∗ (Gr-92), where Q† is the Hermitian conjugate and |αi and |βi are general vectors. c) Is Eop formally a quantum mechanical observable? Does the projection postulate (Gr-105) on the measurement of energy apply to Eop ? HINT: The last question is trickier than it seems. 008 qfull 00040 2 3 0 moderate math: solving an eigenproblem Extra keywords: (Gr-94:3.22) also diagonalizing a matrix. 6. Consider   1 1−i T = . 1+i 0 a) Is T Hermitian? b) Solve for the eigenvalues. Are they real? c) Determine the normalized eigenvectors. Since eigenvectors are not unique to within a phase factor, the marker insists that you arrange your eigenvectors so that the first component of each is 1. Are they orthogonal? d) Using the eigenvectors as columns construct the inverse of a unitary matrix transformation which applied to T gives a diagonalized version of T . Find this diagonalized version T d . What is special about the diagonal elements? e) Compare the determinant det|T |, trace Tr(T ), and eigenvalues of T to those of T d . 008 qfull 00500 2 5 0 moderate thinking: x-op in general formalism Extra keywords: and k-op and p-op in general formalism too 7. The general formalism of quantum mechanics requires states to be vectors in Hilbert spaces and dynamical variables to be governed or determined (choose your verb) by observables (Hermitian operators with complete sets of eigenstates: i.e., sets that form a basis for the Hilbert space). These requirements are a Procrustian bed for the position, wavenumber (or momentum), and kinetic energy operators. These operators have complete sets of eigenvectors in a sense, but those eigenvectors arn’t in any Hilbert space, because they can’t be normalized. Nevertheless everything works out consistently if some identifications are made. The momentum and kinetic energy eigenstates are the same as the wavenumber eigenstates, and so we won’t worry about them. The momentum and kinetic energy eigenvalues are different, of course. NOTE: Procrustes (he who stretches) was a robber (or cannibal) with a remarkable bed that fit all guests—by racking or hacking according to whether small or tall. Theseus fit Procrustes to his own bed—and this was before that unfortunate incident with the Minotaur. a) Consider the xop eigenproblem in the general form xop |xi = x|xi , where x is the eigenvalue and |xi is the eigenvector. The eigenvalues x and eigenvectors |xi’s form continuous, not discrete, sets. The unity operator for the xop basis is therefore Z 1 = dx |xihx| , where it is implied that the integral is over all space. An ideal measurement of position yields x and, by quantum mechanical postulate, puts the system is in state |xi. But the

60 Chapt. 8 Operators, Hermitian Operators, Braket Formalism system can’t be really be in an unnormalizable state which is what the |xi’s turn out to be. The system can be in an integral linear combination of such states. Expand a general state |Ψi in the xop basis and identify what |Ψi is in the position representation. Then identify what the inner product of two xop eigenvectors hx′ |xi must be. Why can’t the |xi be in the Hilbert space? What is the position representation of |xi? Prove that xop in the position operator is just x itself. b) Repeat part (a), mutatis mutandis, for kop . c) What must hx|ki be? This is just an identification, not a proof—there are no proofs. HINT: Expand |Ψi in the wavenumber representation and then operate on |Ψi with hx|.

d) What is hk|k ′ i if we insert the position representation unit operator given the answer to part (c).

e) In order to have consistency with past work what must the matrix elements hx|kop |xi, hx|Hop |xi, and hk|xop |ki be. Note these are just identifications, not proofs—there are no proofs. We omit hk|Hop |ki—you’re not ready for hk|Hop |ki as Jack Nicholson would snarl—if he were teaching intro quantum. 008 qfull 00100 2 5 0 moderate thinking: expectation values two ways Extra keywords: (Gr-108:3.35) 8. Consider the observable Q and the general NORMALIZED vector |Ψi. By quantum mechanics postulate, the expectation of Qn , where n ≥ 0 is some integer, for |Ψi is hQn i = hΨ|Qn |Ψi . a) Assume Q has a discrete spectrum of eigenvalues qi and orthonormal eigenvectors |qi i. It follows from the general probabilistic interpretation postulate of quantum mechanics, that expectation value of Qn for |Ψi is given by hQn i =

X i

qin |hqi |Ψi|2 .

Show that this expression for hQn i also follows from the one in the preamble. What is P 2 i |hqi |Ψi| equal to?

b) Assume Q has a continuous spectrum of eigenvalues q and Dirac-orthonormal eigenvectors |qi. (Dirac-orthonormal means that hq ′ |qi = δ(q ′ − q), where δ(q ′ − q) is the Dirac delta function. The term Dirac-orthonormal is all my own invention: it needed to be.) It follows from the general probabilistic interpretation postulate of quantum mechanics, that expectation value of Qn for |Ψi is given by hQn i =

Z

dq q n |hq|Ψi|2 .

n RShow that 2this expression for hQ i also follows from the one in the preamble. What is dq |hq|Ψi| equal to?

008 qfull 00200 2 5 0 moderate thinking: simple commutator identities 9. Prove the following commutator identities.

a) [A, B] = −[B, A].   X X X b)  ai Ai , bj Bj  = ai bj [Ai , Bj ], where the ai ’s and bj ’s are just complex numbers. i

j

ij

Chapt. 8 Operators, Hermitian Operators, Braket Formalism 61 c) [A, BC] = [A, B]C + B[A, C]. d) [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0. This has always seemed to me to be perfectly useless however true. e) (c[A, B])† = c∗ [B † , A† ], where c is a complex number. f) The special case of the part (e) identity when A and B are Hermitian and c is pure imaginary. Is the operator in this special case Hermitian or anti-Hermitian? 008 qfull 00300 3 5 0 tough thinking: nontrivial commutator identities Extra keywords: (Gr-111:3.41) but considerably extended. 10. Prove the following somewhat more difficult commutator identities. a) Given [B, [A, B]] = 0 ,

[A, F (B)] = [A, B]F ′ (B) ,

prove

where A and B are general operators aside from the given condition and F (B) is a general operator function of B. HINTS: Proof by induction is probably best. Recall that any function of an operator is (or is that should be) expandable in a power series of the operator: i.e., F (B) =

∞ X

fn B n ,

n=0

where fn are constants. b) [x, p] = i− h. c) [x, pn ] = i− h npn−1 . HINT: Recall the part (a) answer. d) [p, xn ] = −i− h nxn−1 . HINT: Recall the part (a) answer. 008 qfull 01300 2 5 0 moderate thinking: general uncertainty principle Extra keywords: Gr-108’s proof 11. In quantum mechanics we believe that states are described by vectors in an abstract Hilbert space. The Hilbert space is an inner product vector space among other things. An inner product vector space (as physicists view it anyway) has vectors written |αi (the kets) and dual space vectors written hα| (the bras). We require that there is some rule such that the inner product operation hα|βi yields a complex number for general vectors |αi and |βi. The other properties required are: hβ|αi = hα|βi∗ , hα|αi ≥ 0 , where the equality holds if and only if |αi = 0 (i.e., |αi is the null vector), and hα|(b|βi + c|γi) = bhα|βi + chαγi . The norm of a vector |αi is defined by ||α|| =

p hα|αi .

The vectors are transformed into new vectors by operators: e.g., |α′ i = Q|αi ,

62 Chapt. 8 Operators, Hermitian Operators, Braket Formalism where Q is some operator. The Schwarz inequality hα|αihβ|βi ≥ |hα|βi|2 is a feature of inner product vector spaces. A concrete interpretation of the Schwarz inequality is that a vector must have a magnitude larger than or equal to any component of the vector along any axis. This can be seen by dividing the Schwarz inequality equality by hα|αi assuming that |αi is not null and interpreting hˆ α|βi as the component of |βi in the |ˆ αi direction where |ˆ αi =

|αi . ||α||

In quantum mechanics, the inner product operation in the 1-dimensional position representation, for example, is defined by hα|βi =

Z

∞

α(x)∗ β(x) dx ,

−∞

where α(x) and β(x) are square integrable functions. The dynamical variables of quantum mechanics are governed by physically relevant Hermitian operators with complete sets of eigenvectors. These operators, not too cogently, are called observables. The mean value or expectation value for an ensemble of identical states |αi of a dynamical variable governed by an observable Q is given by hα|Q|αi . The expectation value can be viewed as the overlap integral between the states |αi and Q|αi. The Hermitian conjugate Q† of an operator Q is defined by hα|Q|βi = hβ|Q† |αi∗ . If Q† = Q, then Q is a Hermitian operator. Hermitian operators have only real expectation values. This follows from the general definition of Hermitian conjugation: if Q† = Q, then hα|Q|αi = hα|Q|αi∗ . This result implies that the eigenvalues of a Hermitian operator are pure real since they are the expectation values of the eigenstates. We will not prove it here, but it is true also that nondegenerate eigenstates of a Hermitian operator are orthogonal and that the set of eigenstates of a Hermitian operator are a complete set guaranteed if the space is finite dimensional (Gr-93). But if the space is infinite dimensional completeness is not guaranteed for the set of eigenstates (Gr-99). The Hermitian operators with complete sets can be quantum mechanical observables. Given the foregoing (and anything I’ve left out), the general uncertainty principle is a necessary consequence. This principle is σA σB ≥

1 |hi[A, B]i| , 2

Chapt. 8 Operators, Hermitian Operators, Braket Formalism 63 where A and B are observables, [A, B] = AB − BA is the commutator of A and B, and σA and σB are standard deviations for observables A and B. The standard deviation squared (i.e., the variance) of A for a state |αi, for example, is given by 2 σA = hα|(A − hAi)2 |αi ,

where hAi = hα|A|αi. Since A is a Hermitian operator, one can write 2 σA = hα|(A − hAi)(A − hAi)|αi = hα|(A − hAi)|f i

= hf |(A − hAi)|αi = hf |f i ,

where we have used the fact that expectation value of a Hermitian operator is real and have defined |f i = (A − hAi)|αi. a) Define |gi = (B − hBi)|αi. Now prove that 2 2 σA σB ≥ |hf |gi2 .

b) Evaluate hf |gi in terms of expectation values of A and B for state |αi. For brevity the state can be left implied where convenient: i.e., hAi stands for hα|A|αi, etc. Remember A and B are Hermitian and hf |gi is in general complex. c) Show that for any complex number z |z|2 = [Re(z)]2 + [Im(z)]2 , where Re(z) =

1 (z + z ∗ ) 2

and

Im(z) =

Re(hf |gi)

and

Im(hf |gi) .

1 (z − z ∗ ) . 2i

Then evaluate

d) In the classical analog case, A and B would just be dynamical variables whose statistical properties can be evaluated for some probability density. What does Re(hf |gi) correspond to classically? e) Now show that the general uncertainty inequality follows from the inequality you found in the part (a) answer. Is there a classical analog to the general uncertainty relation? Why or why not? f) Given that A and B are Hermitian, show that [A, B] is not Hermitian in general, but that i[A, B] is. g) Find the uncertainty relation for operators x and p: i.e., the Heisenberg uncertainty relation. h) What relationship between states |f i and |gi will make the general uncertainty an equality? i) Apply the relationship for equality to the case of x and p and obtain a differential equation for a one-dimensional state for which the Heisenberg uncertainty equality holds. Solve this differential equation for the minimum uncertainty wave packet. What kind of wave packet do you have? In what physical cases is the minimum uncertainty wave packet a solution to the Schr¨odinger equation.

64 Chapt. 8 Operators, Hermitian Operators, Braket Formalism j) Explain why the Heisenberg uncertainty equality is useful for understanding the ground states of many systems. 008 qfull 01400 3 5 0 tough thinking: general uncertainty principle Extra keywords: A dumb proof, and outdated by my current understanding. 12. You have a strange looking operator: ℓ=

δR δQ +i . ∆Q ∆R

where Q and R are general Hermitian operators, ∆Q and ∆R are the standard deviations of Q and R, and δQ ≡ Q − hQi and δR ≡ R − hRi . (a) Write down the Hermitian conjugate ℓ† .

(b) Show ℓ† ℓ a Hermitian operator and that it is a positive definite operator: i.e, that hℓ† ℓi ≥ 0 . HINT: If you have to think about these results for more than a few seconds, then just assume them and go on. (c) Multiply out ℓ† ℓ and gather the cross terms into a commutator operator. Substitute for δQ and δR in the commutator using their definitions and simplify it. (d) Evaluate the expectation value of the multiplied out ℓ† ℓ operator. remembering the definition of standard deviation.

Simplify it

(e) Remembering the positive definite result from part (b), find an inequality satisfied by ∆Q∆R. (f) Since the whole of the foregoing mysterious procedure could have been done with Q and R interchanged in the definition of ℓ, what second inequality must be satisfied by ∆Q∆R. (g) What third ∆Q∆R inequality is implied by two previous ones. 008 qfull 01500 2 3 0 moderate math: x-H uncertainty relation Extra keywords: (Gr-110:3.39) 13. Answer the following questions. a) What is the uncertainty relation for operators x and H? Work it out until the expectation value is for the momentum operator p. b) What is the time-dependent expression for any observable expectation value hQi = hΨ(t)|Q|Ψ(t)i when the state |Ψ(t)i is expanded in the discrete set of stationary states (i.e., energy eigenstates) with their time-dependent factors included to allow for the time dependence of |Ψ(t)i. Let the set of stationary states with explicit time dependence be − {e−iEj t/ h |φj i}. Note functions of t commute with observables: observables may depend on time, but they don’t contain time derivatives. −

c) If state |Ψ(t)i from part (b) is itself the stationary state e−iEj t/ h |φj i, what the expectation value? Is the expectation value time independent? d) Derive the special form of the uncertainty relation for operators x and H for the case of a stationary state of H? What, in fact, is σH for a stationary state? HINT: Remember Ehrenfest’s theorem. 008 qfull 01600 3 5 0 tough thinking: neutrino oscillation Extra keywords: (Gr-120:3.58) 14. There are systems that exist apart from 3-dimensional Euclidean space: they are internal degrees of freedom such intrinsic spin of an electron or the proton-neutron identity of a nucleon (isospin:

Chapt. 8 Operators, Hermitian Operators, Braket Formalism 65 see, e.g., En-162 or Ga-429). Consider such an internal system for which we can only detect two states:     1 0 |+i = and |−i = . 0 1 This internal system is 2-dimensional in the abstract vector sense of dimensional: i.e., it can be described completely by an orthonormal basis of consisting of the 2 vectors we have just given. When we measure this system we force it into one or other of these states: i.e., we make the fundamental perturbation. But the system can exist in a general state of course:   c+ (t) |Ψ(t)i = c+ (t)|+i + c− (t)|−i = . c− (t) a) Given that |Ψ(t)i is NORMALIZED, what equation must the coefficients c+ (t) and c− (t) satisfy. b) For reasons only known to Mother Nature, the states we can measure (eigenvectors of whatever operator they may be) |+i and |−i are NOT eigenstates of the Hamiltonian that governs the time evolution of internal system. Let the Hamiltonian’s eigenstates (i.e., the stationary states) be |+′ i and |−′ i: i.e., H|+′ i = E+ |+′ i

H|−′ i = E− |−′ i ,

and

where E+ and E− are the eigen-energies. Verify that the general state |Ψ(t)i expanded in these energy eigenstates, −

−

|Ψ(t)i = c+ e−iE+ t/ h |+′ i + c− e−iE− t/ h |−′ i satisfies the general vector form of the Schr¨odinger equation: ∂ i− h |Ψ(t)i = H|Ψ(t)i . ∂t HINT: This requires a one-line answer. c) The Hamiltonian for this internal system has no differential operator form since there is no wave function. The matrix form in the |+i and |−i representation is H=



f g

g f



.

Given that H is Hermitian, prove that f and g must be real. d) Solve for the eigenvalues (i.e., eigen-energies) of Hamiltonian H and for its normalized eigenvectors |+′ i and |−′ i in column vector form. e) Given at t = 0 that   a |Ψ(0)i = b show that 1 1 − − |Ψ(t)i = √ (a + b)e−i(f +g)t/ h |+′ i + √ (a − b)e−i(f −g)t/ h |−′ i 2 2 and then show that      cos(gt/− h) −i sin(gt/− h) −if t/− h a |Ψ(t)i = e +b . −i sin(gt/− h) cos(gt/− h)

66 Chapt. 8 Operators, Hermitian Operators, Braket Formalism HINT: Recall the time-zero coefficients of expansion in basis {|φi i} are given by hφi |Ψ(0)i. f) For the state found given the part (e) question, what is the probability at any time t of measuring (i.e., forcing by the fundamental perturbation) the system into state   1 |+i = ? 0 HINT: Note a and b are in general complex. g) Set a = 1 and b = 0 in the probability expression found in the part (f) answer. What is the probability of measuring the system in state |+i? in state |−i? What is the system doing between the two states? NOTE: The weird kind of oscillation between detectable states we have discussed is a simple model of neutrino oscillation. Just as an example, the detectable states could be the electron neutrino and muon neutrino and the particle oscillates between them. Really there are three flavors of neutrinos and a three-way oscillation may occur. There is growing evidence that neutrino oscillation does happen. (This note may be somewhat outdated due to that growth of evidence.) 008 qfull 01700 2 5 0 moderate thinking: operator product rule Extra keywords: Reference Ba-134 15. A function of an operator A can be defined by a power series f (A, λ) =

∞ X

ak (λ)Ak ,

k=0

where λ is an example of c-number parameter of the function and convergence is guaranteed by faith alone. For example, eλA means that eλA =

∞ X λk k=0

k!

Ak .

a) Show that a sensible definition of the derivative of f (A, λ) with respect to λ is ∞

X dak df = Ak . dλ dλ k=0

HINT: Differentiate a general matrix element of f (A, λ). b) Given the definition of the operator derivative of part (a), find product rule for operator derivatives. HINT: Differentiate a general matrix element of f (A, λ)g(B, λ), where f and g are operator functions of general operators A and B. 008 qfull 01800 3 5 0 hard thinking: common eigensets, CSCO Extra keywords: See CDL-139–144 16. One can always construct a common basis (or common eigenset) for commuting observables (i.e., Hermitian operators that allow complete eigensets). Let us investigate this property. a) First, an easy problem. Say you are given obserables A and B and they have a common eigenset {|uia,b i} such that A|uia,b i = a|uia,b i

and

B|uia,b i = b|uia,b i ,

Chapt. 8 Operators, Hermitian Operators, Braket Formalism 67 where a and b are eigenvalues and i labels different states that are degenerate with respect to both eigenvalues: i.e., have the same a and b eigenvalues. Thus specifying a, b, and i fully specifies a state. Show that A and B must commute. b) Now you are given [A, B] = 0 and the eigenset of A {|uia i}, where i labels states that are degenerate with respect to eigenvalues a. Show that you can construct a common eigenset for A and B. HINT: Formally diagonalize B in the set {|uia i} making use of commutation to show that many matrix elements are zero. c) Say {A, B, C, . . .} constitutes a complete set of commuting operators (i.e., a CSCO). What can say about the common eigenset of {A, B, C, . . .}?

Chapt. 9 Time Evolution and the Heisenberg Representation

Multiple-Choice Problems 009 qmult 00100 1 4 1 moderate deducto-memory: constant of the motion 1. What are the conditions for observable Q to be a constant of the motion. a) b) c) d) e)

[H, Q] = 0 [H, Q] 6= 0 [H, Q] > 0 [H, Q] < 0 [H, Q] ≥ 0

and and and and and

∂Q/∂t = 0. ∂Q/∂t 6= 0. ∂Q/∂t > 0. ∂Q/∂t < 0. ∂Q/∂t ≥ 0.

Full-Answer Problems 009 qfull 00100 3 5 0 tough thinking: time evolution, virial theorem Extra keywords: (Gr-117:3.53, Gr2005-126:3.31) No one remembers Dorothy Lamour. 1. Answer the following questions that lead up to the proof of the virial theorem. HINTS: The answers to the earlier parts help answering the later parts. But you can still answer some later parts even if you don’t get all the earlier parts. −

a) Given that e−iEn t/ h |φn i, a STATIONARY STATE (i.e., an eigen-energy state) of a time-independent Hamiltonian with its time-dependence factor explicitly shown, show that the expectation value for this state of any time-independent operator A is a constant with respect to time: i.e., dhAi =0. dt HINT: This is easy. b) Given that |φn i is a stationary state of H and A is a general operator (i.e., it doesn’t have to be an observable or Hermitian), show that hφn |[H, A]|φn i = 0 . HINT: This is not so hard. Recall the formal definition of the Hermitian conjugate of general operator A is hα|A|βi = hβ|A† |αi∗ . c) Prove that [A, BC] = [A, B]C + B[A, C] for general operators A, B, and C. d) Prove [x, p] = i− h. e) Prove that [H, x] = − 68

i− h p. m

Chapt. 9 Time Evolution and the Heisenberg Representation 69 f) Prove that [H, p] = i− h



∂V ∂x



.

g) Starting with the general time evolution equation (or general equation of motion) for general observable A   dhAi 1 ∂A + − hi[H, A]i = dt ∂t h show that

  ∂V dhxpi , = 2hT i − x dt ∂x

where T = p2 /(2m) is the kinetic energy operator. h) Show that dhxpi dhpxi = . dt dt Whe quantity xp is the one-particle, one-dimensional virial operator. In classical physics, it would be a dynamical variable. HINT: This is easy. i) Now for a STATIONARY STATE prove the 1-d virial theorem:   1 ∂V hT istationary = x . 2 ∂x stationary HINT: Don’t forget part (a) and what the general equation of motion says. j) Given potential V (x) ∝ xλ , show that the virial theorem reduces to hT istationary =

λ hV istationary . 2

009 qfull 01000 2 5 0 moderate thinking: Heis. Rep. evolution Extra keywords: See Ba-137 2. Say A is an operator in the Schr¨odinger representation: this is the ordinary representation of beginning quantum mechanics. The Heisenberg representation of this operator for a system with Hamiltonian H is − − A(t) = eiHt/ h Ae−iHt/ h , where exp(−iHt/− h ) is operator function of H defined by e

−iHt/− h

ℓ  ∞ X 1 −iHt = . − ℓ! h ℓ=0

Show definitively that 1 dA(t) = − [A(t), H] + dt ih



∂A ∂t



(t) ,

where the last term accounts for any implicit time dependence in A. HINTS: Take the time derivative of a general matrix element of A(t) with the general states expanded in the eigenkets of H. Note that in the Heisenberg representation states don’t have any time dependence: all the time-dependence is in the operators. 009 qfull 01500 3 5 0 tough thinking: translation operator Extra keywords: (Ba-145:2)

70 Chapt. 9 Time Evolution and the Heisenberg Representation 3. Here are some fun proofs to do that are all in the Schr¨odinger representation, except for the last one. a) Prove

[~r, f (~ p · ~u)] = i− h ∇p~ f (~ p · ~u) ,

where ~u is just a c-number vector (i.e., a vector consisting of ordinary scalar constant components) and the gradient operator is just a specially defined bit of formalism that means differentiate with respect to components of p~ to form a gradient as if they were ordinary variables. HINT: Recall f (~ p · ~u) =

∞ X ℓ=0

aℓ (~ p · ~u)ℓ ,

where the aℓ are some scalar coefficients and convergence is assumed. b) Using the part (a) result, prove that −

−

ei~p·~u/ h ~re−i~p·~u/ h = ~r + ~u . c) Now prove that −

|Ψ′ i = e−i~p·~u/ h |Ψi is the same state as |Ψi translated by ~u. In the position representation −

Ψ(~r )′ = e−i~p·~u/ h Ψ(~r ) = Ψ(~r − ~u) . d) Given −

|Ψ′ i = e−i~p·~u/ h |Ψi , where |Ψ′ i and |Ψi are in the Heisenberg representation, show that |Ψ′ (t)i (the state |Ψ′ i in the Schr¨odinger representation) evolves according to −

|Ψ′ (t)i = e−i~p(−t)·~u/ h |Ψ(t)i , where p~(−t) is the momentum operator in the Heisenberg representation at −t.

Chapt. 10 Measurement

Multiple-Choice Problems 010 qmult 00100 1 1 1 easy memory: fundamental perturbation 1. In an ideal quantum mechanical measurement of an observable A: a) the measurement always detects an EIGENVALUE of the observable and projects the system into an EIGENSTATE of the observable corresponding to that eigenvalue. b) the measurement always detects an EXPECTATION VALUE of the observable and projects the system into an EIGENSTATE of the observable. c) the measurement always detects an EXPECTATION VALUE of the observable and projects the system into an NON-EIGENSTATE of the observable. d) the measurement always detects an 3 EIGENVALUES of the observable and projects the system into an NON-EIGENSTATE of the observable. e) The measurement always detects an EXPECTATION VALUE of the observable and projects the system into a STATIONARY STATE.

Full-Answer Problems

71

Chapt. 11 The Central Force Problem and Orbital Angular Momentum

Multiple-Choice Problems 011 qmult 00100 1 4 3 easy deducto-memory: central-force 1. In a central-force problem, the magnitude of central force depends only on: a) b) c) d) e)

the the the the the

angle of the particle. vector ~r from the center to the particle. radial distance r from the center to the particle. magnetic quantum number of the particle. uncertainty principle.

011 qmult 00200 1 1 2 easy memory: separation of variables 2. The usual approach to getting the eigenfunctions of the Hamiltonian in multi-dimensions is: a) non-separation of variables. b) separation of variables. c) separation of invariables. d) non-separation of invariables. e) non-separation of variables/invariables. 011 qmult 00210 1 1 3 easy memory: separation of variables 3. Say you have a differential equation of two independent variables x and y and you want to look for solutions that can be factorized thusly f (x, y) = g(x)h(y). Say then it is possible to reorder equation into the form LHS(x) = RHS(y) , where LHS stands for left-hand side and RHS for right-hand side. Well LHS is explicitly independent of y and implicitly independent of x: ∂LHS =0 ∂y

∂LHS ∂RHS = =0. ∂x ∂x

and

Thus, LHS is equal to a constant C and necessarily RHS is equal to the same constant C which is called the constant of separation (e.g., Arf-383). The solutions for g(x) and h(y) can be found separately and are related to each other through C. The solutions for f (x, y) that cannot be factorized are not obtained, of course, by the described procedured. However, if one obtains complete sets of g(x) and h(y) solutions for the x-y region of interest, then any solution f (x, y) can be constructed at least to within some approximation (Arf-443). Thus, the generalization of the described procedure is very general and powerful. It is called: a) separation of the left- and right-hand sides. b) partitioning. c) separation of the variables. d) solution factorization. e) the King Lear method. 011 qmult 00300 1 4 2 easy deducto-memory: relative/cm reduction 4. “Let’s play Jeopardy! For $100, the answer is: By writing the two-body Schr¨odinger equation in relative/center-of-mass coordinates.” How do you

, Alex?

a) reduce a ONE-BODY problem to a TWO-BODY problem b) reduce a TWO-BODY problem to a ONE-BODY problem 72

Chapt. 11 The Central Force Problem and Orbital Angular Momentum 73 c) solve a one-dimensional infinite square well problem d) solve for the simple harmonic oscillator eigenvalues e) reduce a TWO-BODY problem to a TWO-BODY problem 011 qmult 00310 1 4 4 easy deducto-memory: reduced mass 5. The formula for the reduced mass m for two-body system (with bodies labeled 1 and 2) is: a) m = m1 m2 . e) m =

1 . m1

b) m =

1 . m1 m2

c) m =

m1 + m2 . m1 m2

d) m =

m1 m2 . m1 + m2

011 qmult 00400 1 4 2 easy deducto memory: spherical harmonics 1 6. The eigensolutions of the angular part of the Hamiltonian for the central force problem are the: a) linear harmonics. b) spherical harmonics. c) square harmonics. d) Pythagorean harmonics. e) Galilean harmonics. 011 qmult 00410 1 4 4 easy deducto-memory: spherical harmonics 2 Extra keywords: mathematical physics 7. “Let’s play Jeopardy! For $100, the answer is: They form a basis or complete set for the 2dimensional space of the surface a sphere which is usually described by the angular coordinates of spherical polar coordinates.” What are the

, Alex?

a) Hermite polynomials b) Laguerre polynomials c) associated Laguerre polynomials e) Chebyshev polynomials

d) spherical harmonics

011 qmult 00420 1 4 3 easy deducto memory: spherical harmonic Y00 8. Just about the only spherical harmonic that people remember—and they really should remember it too—is Y00 =: a) eimφ .

b) r2 .

1 c) √ . 4π

d) θ2 .

e) 2a−3/2 e−r/a .

011 qmult 00500 1 4 2 easy deducto-memory: spdf designations 9. Conventionally, the spherical harmonic eigenstates for angular momentum quantum numbers ℓ = 0, 1, 2, 3, 4, ... are designated by: a) b) c) d) e)

a, b, c, d, e, etc. s, p, d, f , and then alphabetically following f : i.e., g, h, etc. x, y, z, xx, yy, zz, xxx, etc. A, C, B, D, E, etc. $@%&*!!

011 qmult 00510 1 4 3 easy deducto-memory: s electrons 10. “Let’s play Jeopardy! For $100, the answer is: What the ℓ = 0 electrons (or zero orbital angular momentum electrons) are called in spectroscopic notation.” What are

, Alex?

a) the Hermitian conjugates e) h electrons

b) Herman’s Hermits

c) s electrons

d) p electrons

74 Chapt. 11 The Central Force Problem and Orbital Angular Momentum

Full-Answer Problems 011 qfull 00100 2 5 0 moderate thinking: 2-body reduced to 1-body problem Extra keywords: (Gr-178:5.1) 1. The 2-body time-independent Schr¨odinger equation is −

2 2 − − h h ∇21 ψ − ∇2 ψ + V ψ = Etotal ψ . 2m1 2m2 2

If the V depends only on ~r = ~r2 − r1 (the relative vector), then the problem can be separate into two problems: a relative problem 1-body equivalent problem and a center-of-mass 1-body equivalent problem. The center of mass vector is ~ = m1~r1 + m2~r2 , R M where M = m1 + m2 . ~ and ~r. a) Determine the expressions for ~r1 and ~r2 in terms of R b) Determine the expressions for ∇21 and ∇22 in terms of ∇2cm (the center-of-mass Laplacian operator) and ∇2 (the relative Laplacian operator). Then re-express kinetic operator 2 2 − − h h 2 ∇ − ∇2 − 2m1 1 2m2 2

in terms of ∇2cm and ∇2 . HINTS: The x, y, and z direction components of vectors can ~ and ~r) (i.e., X and x) all be treated separately and identically since x components of R depend only on x1 and x2 , etc. You can introduce a reduced mass to make the transformed kinetic energy operator simpler. c) Now separate the 2-body Schr¨odinger equation assuming V = V (~r ). What are the solutions of the center-of-mass problem? How would you interpret the solutions of the relative problem? HINT: I’m only looking for a short answer to the interpretation question. 011 qfull 00200 2 3 0 moderate math: central-force azimuthal component solution Extra keywords: solving the azimuthal component of the central force problem 2. In the central force problem, the separated azimuthal part of the Schr¨odinger equation is: d2 Φ = −m2ℓ Φ , dφ2 where −m2ℓ is the constant of separation for the azimuthal part. The constant has been parameterized in terms of mℓ (which is not mass) since it turns out that for normalizable (and therefore physically allowed) solutions that m must be an integer. The mℓ quantity is the z-component angular momentum quantum number or magnetic quantum number (MEL-59; ER-240). The latter name arises since the z-components of the angular momentum manifest themselves most noticeably in magnetic field phenomena. a) Since the differential equation is second order, there should should be two independent solutions for each value of m2ℓ . Solve for the general solution Φ for each m2ℓ : i.e., the solution that is a linear combination of the two independent solutions with undetermined coefficients. Note that writing the separation constant as m2ℓ is so far just a parameterization and nothing yet demands that m2ℓ be greater than zero or pure real. HINT: Use an exponential

Chapt. 11 The Central Force Problem and Orbital Angular Momentum 75 trial function with exponent ±(a + ib) with a and b real. Also remember the special case of m2ℓ = 0. b) The solutions are continuous and so that quantum mechanical requirement is met. But another one must be imposed for the azimuthal coordinate: the single-valuedness condition. Since we have no interpretation for multi-valuedness, we micropostulate that it doesn’t happen. Impose the single-valuedness condition on the generl solution Φ = Ae(a+ib)φ + Be−(a+ib)φ , and show that a = 0 and mℓ must be an integer. Remember to consider the special case where mℓ = 0? c) What are the eigenfunction solutions for the z-component of the angular momentum operator − h ∂ Lz = . i ∂φ What are the eigenvalues that satisfy single-valuedness and continuity? What is the relationship between these eigenfunction solutions and the azimuthal angle part of the hydrogenic atom wave functions? d) Normalize the allowed eigensolutions of Lz Note these solutions are, in fact, conventionally left unnormalized: i.e., the coefficient of the special function that is the solution is left as just 1. Normalization is conventionally imposed on the total orbital angular momentum solutions, spherical harmonics. 011 qfull 01000 3 5 0 tough thinking: the nearly rigid rotator 3. You have a 3-dimensional system consisting of two non-identical particles of masses m1 and m2 . The two particles form a nearly rigid rotator. The relative time-independent Schr¨odinger equation for the system is: #   2 − L2 h 1 ∂ 2 ∂ r + + V (r) Ψ(r, θ, φ) = EΨ(r, θ, φ) , − 2µ r2 ∂r ∂r 2µr2

"

where r, θ, and φ are the relative coordinates, µ = m1 m2 /(m1 + m2 ) is the reduced mass, and the potential is  0, for r ∈ [a − ∆a/2, a + ∆a/2]; V (r) = ∞ , otherwise. a) Assume that ∆a is so much smaller than a that L2 /(2µr2 ) ≈ L2 /(2µa2 ). Now separate the equation into radial and angular parts using Erad and Erot as the respective separation constants: Erad + Erot = E. Let the radial solutions be R(r). You know what the angular solutions should be. Write down the separated equations. b) For the radial equation assume that r varies so much more slowly than R over the region of non-infinite potential that   1 ∂ ∂2R 2 ∂R r ≈ 2 r ∂r ∂r ∂r2 in that region. Change the coordinate variable to x = r−(a−∆a/2) for simplicity: the noninfinite region of the potential then is then the x range [0, ∆a]. With this approximation solve for the radial eigenstates and eigen-energies. Normalize the eigenstates. HINTS: Holy d´ej` a vue all over again Batman, it’s the 1-dimensional infinite square well problem. Don’t mix up a and ∆a.

76 Chapt. 11 The Central Force Problem and Orbital Angular Momentum c) Write down the eigenstates (just their general symbol, not expressions) and eigen-energy expression for the rotational equation. What is the degeneracy of each eigen-energy? HINTS: You shouldn’t being trying to solve the equation. You should know what the eigenstates are. d) Write the general expression for the total wave function. How many quantum numbers does it depend on? e) Write down the general expression for the total energy. Which causes a greater change in energy: a change of 1 in the quantum number controling the radial energy or a change of 1 in the quantum number controling the rotational energy? Remember ∆a 0, but that k/(µω 2 ) > 1. Give a schematic energy-level diagram in this case. e) Now assume that the two particles are identical spin-0 bosons. Note that identical means they now have the same mass. Given the symmetry requirement for boson states, which solutions (specified by the nCM and nREL quantum numbers) are not physically allowed? f) Now assume that the two particles are identical spin-1/2 fermions. Note again that identical means they now have the same mass. But also note they arn’t electrons. Their interactions are determined by the given Hamiltonian only. Because the particles are spin-1/2 fermions, the eigen wave functions for system must be multiplied by appropriate eigen-spinors to specify the full eigenstate. Given the antisymmetry requirement for fermion states, what restrictions are put on the wave function and spinor quantum numbers of an eigenstate? 019 qfull 02600 1 5 0 easy thinking: symmetrization, Slater determinant Extra keywords: (Gr-187:5.7) 9. Say that you solve a Schr¨odinger equation for N identical particles to get the normalized wave function ψ(~r1 , ~r2 , ~r3 , . . . , ~r N ). How would you symmetrize the wave function for bosons? Then how would you symmetrize for fermions all in the spin-up state so that you don’t have spinors to complicate the question? How would you normalize the wave function? 019 qfull 02700 1 5 0 easy thinking: doubly excite He decay Extra keywords: (Gr188.58a) 10. Say you put two electrons into the n = 2 principle quantum number shell of a neutral helium atom and immediately one electron is ejected and the other decays to the ground of the He+ ion. What approximately is the kinetic energy of the ejected electron. NOTE: Without a detailed specification of the doubly-excited helium atom we cannot know exactly what the energies of the excited electrons are. There are two simple approximate choices for their energies: 1) assume that the energy levels of the singly-excited helium atom apply (see, e.g., Gr-189); 2) assume that the Z = 2 hydrogenic energy levels apply. The first choice is probably most in error because it assumes too much electron-electron interaction: the electrons may further apart in the actual doubly-excited state; but, in fact, where they are depends on exactly what doubly excited state they are in. The 2nd choice is certainly wrong by assuming zero electron-electron interaction. 019 qfull 02900 2 5 0 moderate thinking: helium with bosons Extra keywords: (Gr-188:5.9) 11. Describe qualitatively how the helium atom energy level diagram would plausibly change under the following conditions. a) Say the electrons were spin zero bosons. b) Say the electrons were spin 1/2 bosons—a contradiction in postulates, but for the sake of argument have it so. c) Say the electrons were spin 1/2 fermions, but were quantum mechanically distinguishable particles. HINT: In this case the answer is going to be pretty much indefinite. 019 qfull 03000 2 5 0 moderate thinking: Bose-Einstein counting Extra keywords: See Po-13 and Po-47 12. In statistical mechanics, the symmetrization requirement on identical bosons enters in the way that probabilities are assigned to the global states they can form. We will investigate how symmetrization manifests itself in this case.

Chapt. 19 Symmetrization Principle 127 a) Say you had g single-particle states and n distinct particles. How many distinct global states can you form? What is the probability of each global state assuming that each has equal probability? b) Now a trickier case. Say you had g single-particle states and n identical particles. The probability pi that a particle P goes into single-particle state i is INDEPENDENT of what g the other particles do: note i=1 pi = 1, of course. You can construct all possible global states by inserting one particle at a time into the system—can you imagine a global state that cannot be so constructed? Say you do insert the n particles one at a time to the system. The probability of an n-particle global state formed by the insertion sequence ijk . . . ℓ is pi pj pk . . . pℓ which has n factors, of course. But because the particles are identical, each (distinct) global state can be constructed in general by multiple insertion sequences. How many distinct insertion sequences for n particles correspond to a single global state with occupation number set {ni }? If all the pi are equal, what is the probability of a global state with occupation number set {ni } formed by random insertion of particles? The sum of the probabilities for all insertion sequences is 1. Why must this be so on general grounds? Now prove more explicitly that the sum of all inserttion sequence probabilities is 1. HINT: Consider 1=

g X i=1

pi

!n

and a proof by induction. c) Now in the part (b) answer, we didn’t find out how many distinct global states there were. To find this out you need a different counting procedure. Let’s consider finding all possible global states given the following conditions. Imagine that all n particles were distinct and that the order in which you choose the single-particle states to slot them into also matters. To start with you must select a state: you can’t put a particle in a non-state. Then proceed selecting a particle for the current state or a new state until you are out of particles and states. Now did the order of the states matter or the order of the choice of particles? d) Now for classical, non-interacting particles randomly slotted into single-particle states, the probability of each global state is as determined in part (b). Quantum mechanical non-interacting bosons do not act like classical particles. Because of the symmetrization principle—in a way the instructor has never found out—each distinct global state has equal probability. What is this probability for n bosons in g single-particle states? Say that we have all n bosons in one single-particle state. What is the classical probability of this global state? Which is larger the classical probability or the boson probability? What does the last result suggest about the random distributions of bosons relative to classical random distributions? e) Consider two identical coins—say quarters. How many distinct global physical states can be made given that the single-coin states are head and tail? Now toss them up together in a completely randomizing way 36 times. Count the number of distinct global states of each kind that you get? Do the probabilities of each distinct global state appear to be classically random or boson random?

Chapt. 20 Atoms

Multiple-Choice Problems 020 qmult 00100 1 1 1 easy memory: atom defined 1. An atom is a stable bound system of electrons and: a) b) c) d) e)

a single nucleus. two nuclei. three nuclei. a single quark. two quarks.

020 qmult 01000 1 4 1 easy deducto-memory: central potential 2. “Let’s play Jeopardy! For $100, the answer is: The overwhelmingly favored way to solve for the electronic structure of atoms in quantum mechanics.” a) b) c) d) e)

What What What What What

is is is is is

the the the the the

central potential approximation, Alex? non-central potential approximation, Alex? grand central approximation, Alex? atom-approximated-as-molecule method, Alex? electrons-as-bosons approximation, Alex?

Full-Answer Problems 020 qfull 00100 1 5 0 easy thinking: spectrum of He II Extra keywords: (Gr-188:58b) 1. Describe the spectrum of He II (i.e., singly-ionized helium or He+ ) sans perturbations.. 020 qfull 00200 3 3 0 tough math: helium atom 1st order perturbation Extra keywords: (Gr-188:5.10) 2. If one neglects the electron-electron interaction of the helium atom then the spatial ground state is just the product of two hydrogenic states: ψ(~r1 , ~r 2 ) = ψ100 (~r 1 )ψ100 (~r 2 ) =

8 −2(r1 +r2 )/a 1 e , e−2(r1 +r2 )/a = πa3He /8 πa3

where aHe = a/Z = a/2 is the helium Bohr radius and a is the standard Bohr radius (see, e.g., Gr-137–138 and Gr-187). The 1st order perturbation correction to the helium atom ground state is given by hH ′ i , where H ′ is the perturbation Hamiltonian: i.e.,

1 e2 4πǫ0 |~r1 − ~r2 | 128

Chapt. 20 Atoms 129 in MKS units (see, e.g., Gr-187) or e2 |~r1 − ~r2 | in Gaussian CGS units. Note that we use e for both fundamental charge unit and the exponential factor: this is conventional of course: context must decide which is which. a) Analytically calculate  HINTS: Set |~r1 − ~r2 | =

1 |~r1 − ~r2 |



.

q r12 + r22 − 2r1 r2 µ ,

where µ = cos θ is the angle cosine between the vectors. Integrate over all ~r2 space first taking ~r1 as the z axis for spherical coordinates. It helps to switch to dimensionaless variables earlier on. There are no specially difficulties or tricks: just a moderate number of steps that have to be done with tedious care. b) Now the expression for a hydrogenic energy level, sans perturbations, is Z 2 Eryd 13.6Z 2 Z2 1 ≈ − eV , En = − me c2 α2 2 = − 2 n n2 n2 where me is the electron mass, α is the fine structure constant, Z is the nuclear charge, and Eryd ≈ 13.6 eV is the Rydberg energy (see, e.g., Ga-197). In Gaussian CGS units e2 α= − hc

and

a=

− h me cα

(e.g., Ga-199). What is the energy of the helium atom ground state in terms of Rydberg energies and eVs? 020 qfull 01000 2 5 0 moderate thinking: quantum defects Extra keywords: (MEL-220:9.1) Needs some more work, particularly (b) 3. Excited states of atoms can usually be approximated as merely promoting a valence electron to a single-particle state at a higher energy than any of the ones used in the ground state. For high energy single-particle states one often finds that their energies form a quasi-Rydberg series: i.e., Enℓ ≈ −

ERyd , (n − µnℓ )2

where n is the principal quantum number, ℓ is the angular momentum quantum number, ERyd = 13.606 eV is the Rydberg energy, and µnℓ is the quantum defect. I suppose quantum defect gets its name since it accounts for a “defect” in the quantum number. When quantum defects are small the wave functions will be quasi-hydrogen-like and hydrogen-like approximations can be used with some confidence—which is often an immense simplification. (Note I use hydrogen-like, not hydrogenic: hydrogen-like implies that the central potential is like e2 /r, whereas hydrogenic implies the central potential is like ZN e2 /r—at least that’s the way it is in this question. Various uses can be made of the quantum defect parameterization of energy (e.g., Mi-97). In understanding quantum defects, three facts are useful to know. First, the single-particle potential well outside of the core approximates V (r) = −

e2 r

130 Chapt. 20 Atoms for a neutral atom. Second, near the nucleus of the atom, the single-particle potential approximates the bare nucleus potential V (r) = −

ZN e2 , r

where ZN is the nuclear charge. Third, wave functions in central potentials for small radius tend to go as rℓ . This result is rather robust since it is true of hydrogenic wave functions (MEL-57; Ar-622) and spherical Bessel functions (Ha-37) which are the radial solutions of the infinite spherical well. Only s states (i.e., ℓ = 0 states) have non-zero probability amplitude at r = 0. a) Find the formula for the quantum defect in terms of the energy of the energy level. Then compute the quantum defects for sodium and lithium given the following table. HINT: A computer program would lessen the labor. Table: Observed Energy Levels in Electronvolts (not to modern accuracy) n

s

p

d

f

2 3 4 5

−5.3923 −2.0188 −1.0509 −0.6432

−3.5442 −1.5577 −0.8701 −0.5544

... −1.5133 −0.8511 −0.5446

... ... −0.8502 −0.5434

3 4 5 6

−5.1397 −1.9480 −1.0229 −0.6297

−3.0359 −1.3863 −0.7946 −0.5150

−1.5223 −0.8557 −0.5472 −0.3797

... −0.8507 −0.5445 −0.3772

Li

Na

b) As you can see from your table of quantum defects, the quantum defects are NOT nearly zero in general as they would be if the states were almost exactly hydrogen-like at their principal quantum number. Also the quantum defects are NOT just equal to the principal quantum number of highest core shell ncore : i.e., they are not 1 for Li and 2 for Na. Thus, the quantum defect shows that the outer states are not hydrogen-like either for their actual principal quantum number n nor at an effective principal quantum number n − ncore . Why are these two a priori guesses for quantum defects wrong even though the potential is close to that of hydrogen with ZN = 1 in the region where the bulk of the probability for outer states is located. c) Give a reason why quantum defects should have positive values if they are significantly large. Why might small quantum defects be negative? d) Why do quantum defects decrease with ℓ? e) Explain why the Na quantum defects tend to be larger than the Li quantum defects? f) Give a reason why quantum defects may not vanish as n → ∞.

Chapt. 21 Molecules

Multiple-Choice Problems 021 qmult 00100 1 1 1 easy memory: molecule defined 1. A stable bound system of electrons and more than one nucleus with some specific recipe for the numbers of each kind of nuclei is: a) b) c) d) e)

a molecule. an atom. a nucleon. a fullerene. a baryon.

021 qmult 00100 1 4 1 easy deducto-memory: atoms bound in molecules 2. “Let’s play Jeopardy! For $100, the answer is: Because they are formed by bonding atoms and dissociate into atoms?” a) b) c) d) e)

What What What What What

is one good reason for thinking of molecules as bound atoms, Alex? are three bad reasons for thinking of molecules as bound atoms, Alex? is no reason at all for thinking of molecules as bound atoms? is ambiguous answer, Alex? is am-Piguous answer, Alex?

021 qmult 00300 2 1 2 easy memory: molecular energy scales 3. Given electron mass m and typical nuclei mass M , the ratio of electronic, vibrational, and rotational energies for a molecule is of order: a) b) c) d) e)

1 : 1 : 1. 1 : (m/M )1/2 : (m/M ). 1 : (m/M ) : (m/M ). 1 : (m/M )1/4 : (m/M )1/2 . 1 : (m/M )1/4 : (m/M )1/3 .

021 qmult 00400 1 4 4 easy deducto-memory: Born-Oppenheimer approx. Extra keywords: (Ba-473) 4. “Let’s play Jeopardy! For $100, the answer is: This approximation allows one to treat the nuclei in atoms as though they interacted only with an effective potential constructed from actual potentials and the electronic kinetic energy (i.e., the total electronic energy).” a) b) c) d) e)

What What What What What

is is is is is

the the the the the

Alpher-Behte-Gamow recipe, Alex? Einstein-Woody-Allen approximation, Alex? linear combination of atomic orbitals method, Alex? Born-Oppenheimer approximation, Alex? tight-binding approximation, Alex?

021 qmult 00500 1 4 5 easy deducto-memory: tight-binding theory 131

132 Chapt. 21 Molecules 5. “Let’s play Jeopardy! For $100, the answer is: It posits that overlapping wave functions of bound atoms can be treated to some degree in terms of the orbitals of isolated atoms.” a) b) c) d) e)

What What What What What

is is is is is

the genetic algorithm method, Alex? the linear variational method, Alex? tight-binder theory, Alex? tight-wadding theory, Alex? tight-binding theory, Alex?

021 qmult 00600 1 1 3 easy memory: LCAO 6. In LCAO (linear combinations of atomic orbitals method) one uses atomic orbitals as a nonorthonormalized basis set for constructing: a) b) c) d) e)

bound states of nucleons. bound states of photons. inter-atomic bonding and anti-bonding states. intra-atomic stationary states. property-law violating beach-front homes in California.

Full-Answer Problems 021 qfull 00100 2 5 0 moderate thinking: universal sp-coupling parameters Extra keywords: See Ha-95 1. Harrison (Ha-95) presents “universal” sp-bond matrix elements or coupling pararameters:

hs1 |H|s2 i = Vssσ hpz1 |H|pz2 i = Vppσ hs1 |H|pz2 i = Vspσ

2 h π2 − , =− 8 md2 2 3π 2 − h = , 8 md2 2 π − h = , 2 md2

and 2

hpx1 |H|px2 i = hpy1 |H|py2 i = Vppπ = −

h π2 − = Vssσ , 8 md2

where 1 and 2 denote two atoms aligned along the z-axis, H is the single-particle Hamiltonian owing to the cores of the two atoms, the |s1 i, etc., are single-particle atomic orbitals (radial parts of some sort times the cubical harmonics for the angular parts) oriented relative to a common set of axes, d is the inter-nuclei separation, and Vspσ has a π, not π 2 . The s and p subscripts on the V ’s indicate the atomic orbitals in the coupling and the σ and π indicate the indicates the quantum numbers m2 of L2z operator of the molecular orbitals that result from the coupling of the different states: σ is for m2 = 0 and π is for m2 = 1. The reason for the complication of using the eigenvalues of the L2z operator rather than the Lz operator is that the |px i and |py i cubical harmonics are eigenstates of L2z , but not of Lz . Recall the lowest quantum number spherical harmonics Yℓ,m are Y0,0

1 , = √ 4π

Y1,0 =

r

3 cos(θ) , 4π

and

Y1,±1 =

r

3 sin(θ)e±imφ , 8π

Chapt. 21 Molecules 133 where ℓ is the L2 quantum number, m is the Lz quantum number, θ is the angle from the z-axis, and φ is the azimuthal angle. The cubical harmonics are defined by 1 , |si = Y0,0 = √ 4π Y1,1 + Y1,−1 √ |px i = = 2 Y1,1 − Y1,−1 √ |py i = = i 2 and |pz i = Y1,0 =

r

r

r

3 x , 4π r 3 y , 4π r

3 z . 4π r

a) p Verify that polar plots of the “p” cubical harmonics are a touching pair of spheres of radius 3/(4π)/2. (I mean, of course, when you consider the plots as spherical polar coordinate plots.) For |pz i, for example, the spheres touch at the z-axis origin and are aligned with the z-axis: the upper sphere is “positive” and the lower sphere is “negative”: i.e., the radial position from the origin comes out a negative number: one just plots the magnitude. HINTS: It is sufficient to do the proof for |pz i, since the others are the same mutatis mutandis. A diagram would help. b) Interpret the physical significance of the polar plots of the cubical harmonics. c) Show that |px i and |py i are eigenstates of L2z , but not Lz . What other angular momentum operators are they eigenstates of? HINT: Recall Lz =

− h ∂ . i ∂φ

d) Now we come to the question yours truly wanted to ask before chronic digression set in. Write sp-bond coupling parameters in terms of fiducial values in units eV-˚ A2 : e.g., C , d2◦ A

where C is a numerical constant (i.e., an actual value) in eV-˚ A2 and dA◦ is mean nuclei ◦ separation in Angstroms. Then evaluate the constants for dA = 3. HINT: Recall 2 − h = 7.62 eV-˚ A2 , m

where m is the electron mass. 021 qfull 00300 2 3 0 easy math: Li2 with spin Extra keywords: Reference Ha-72 2. Let us consider the single-particle bonding and antibonding states and their energies for Li2 . We assume that the single-particle Hamiltonian of the Li2 molecule is 2 2 −− h 2 −− h 2 H= ∇ + V (~r ) = ∇ + V (~r − ~r1 ) + V (~r − ~r2 ) . 2m 2m

where ~r is measured from the midpoint between the nuclei, ~r1 is the position of nucleus 1, and ~r2 the position of nucleus 2. We are assuming the nuclei are at fixed positions which is the

134 Chapt. 21 Molecules crudest Born-Oppenheimer approximation. The observed equilibrium separation of the nuclei is d = 2.67 ˚ A: we take this to be the fixed separation. We will make the LCAO (linear combination of atomic orbitals) approximation. a) Use the linear variational method to calculate the the bonding and antibonding states and their energies. The basis states are the two atomic orbital 2s states of the atoms: call them |1i and |2i. We assume |1i and |2i are knowns: they have the same energy εs = −5.34 eV. Assume the basis states are orthogonal: a poor approximation actually, but this problem is intended to be heuristic. Make a reasonable approximation to evaluate the diagonal matrix elements. For off-diagonal or coupling matrix elements use Vssσ

2 π2 − h 9.40 eV-˚ A =− = − . 8 md2 d2◦ A

this is one of Harrison’s “universal” sp-bond couplings (Ha-95). Which state is the bonding state and which the antibonding state and why are they so called? b) What is the component of orbital angular momentum of single-particle states about the inter-nuclear axis? How do you know this? What symbol represents this orbital angular momentum component value for molecules? HINT: Recall that the z-component orbital angular momentum operator is − h ∂ , Lz = i ∂φ where φ is the azimuthal angle about the z-axis (MEL-23). c) Now construct six plausible symmetrized two-particle states including spinor from our bonding and antibonding position states: a ground state and the five excited states. So everyone is on the same wavelength let α=

  1 0

and

β=

  0 . 1

What are the approximate energies of these states? Can we construct any more states from the bonding and antibonding states? We, of course, are assuming that there is no spin operator in the Hamiltonian. NOTE: These states may not be very realistic: this is just an exercise. 021 qfull 03000 2 5 0 moderate thinking: molecular relative coordinates Extra keywords: A misconcieved problem. 3. One usually wishes to separate the center of mass and relative parts of the nuclei part of a molecular wave function. For two nuclei, the situation is a two-body problem and can be treated like hydrogenic systems (Da-334). For the general multiple nuclei case, a different approach is needed that treats all nuclei on the same footing. Let ~ri be nucleus i’s position relative to ~ the molecular center of an external inertial frame. Let ~ri′ be nucleus i’s position relative to R mass. We make the approximation that the electrons can be neglected in evaluating the center of mass. NOTE: I was fooled into thinking there was neat way of doing this. The cross term doesn’t vanish. But one probably has to construct independent coordinates in a special way for each kind of molecule. But maybe something is salvageable so I’ll leave this around for now. ~ and ~r′ in terms of the positions ~ri . a) Express R i b) Now express the operators ∂/∂xi and ∂ 2 /∂x2i in terms of operators ∂/∂x′i and ∂/∂X. For simplicity we only consider the x-components of the various coordinates. The y- and z-components are handled similarly.

Chapt. 21 Molecules 135 c) Now show that H=

X p2 X p′2 p2 i i + V ({ri }) = cm + + V ({ri′ }) , 2Mi 2Mtot 2Mi′ i i

where H is the Hamiltonian of the nuclei in the effective potential V ({ri }) (the curly brackets mean “set of”) with no external potential present, p′i are the relative coordinate momentum operators, and Mi Mi′ = 2 . [1 − (Mi /Mtot )]

Chapt. 22 Solids

Multiple-Choice Problems 022 qmult 00100 1 1 3 easy memory: simplest quantum mechanical solid model 1. The simplest quantum mechanical solid model is arguably: a) the hydrogen atom. b) the helium atom. c) the free electron gas model. d) the infinite periodic potential model. e) the finite periodic potential model. 022 qmult 00110 1 1 1 easy memory: infinite square boundary conditions 2. For the free electron gas model of a solid, one common simple choice of boundary conditions is: a) infinite square well boundary conditions. b) finite square well boundary conditions. c) Gaussian well boundary conditions. d) hydrogen atom boundary conditions. e) helium atom boundary conditions. 022 qmult 00130 1 4 4 easy deducto-memory: periodic boundary conditions Extra keywords: mathematical physics 3. “Let’s play Jeopardy! For $100, the answer is: These quantum mechanical boundary conditions for solids, also known a Born-von-Karman boundary conditions, are not realistic in most cases. They are realistic in some cases. For example, for the dimension of a solid that forms a closed loop: e.g., a solid that has donut shape can be have an angular coordinate that must periodic by symmetry over the range [0◦ , 360◦ ]. But whether realistic or not, it can be shown—but no one ever says where—that they lead to the same average behavior as realistic boundary conditions for macroscopically large solid samples. Why are these boundary conditions used at all? Well for one thing they are an ideal kind of boundary conditions that are completely independent of what the surface behavior of solid is. Thus, they are neutral case. For another thing they are easy to use in developments in a particular when dealing with periodic potentials in a solid.” What are

, Alex?

a) infinite square well boundary conditions b) aperiodic boundary conditions c) Rabi-Schwinger-Baym-Sutherland boundary conditions d) periodic boundary conditions e) relaxed boundary conditions 022 qmult 00500 1 4 5 easy deducto-memory: Bloch’s theorem Extra keywords: mathematical physics 4. “Let’s play Jeopardy! For $100, the answer is: It is a theorem in quantum mechanics that applies to systems with periodic potentials. In one dimension, say one has the periodic potential V (x) = V (x + a) where a is the period distance The theorem then says that the wave function must satisfy ψ(x + a) = eKa ψ(x)

or

ψ(x) = eKa ψ(x − a)

or ψ(x + na) = eKna ψ(x)

or 136

ψ(x) = eKna ψ(x − na) ,

Chapt. 22 Solids 137 where K is wave number quantity and n is any integer. Say that one has ψ(x) for the range [0, a]. Then the ψ(x) for the range [na, (n + 1)a] can be evaluated from the last formula: e.g., ψ(na) = eKna ψ(0) What is

and

ψ[n + 1)a] = eKna ψ(a) .”

, Alex?

a) Lorentz’s theorem b) Einstein’s rule c) Schr¨odinger’s last theorem d) Born’s periodicity law e) Bloch’s theorem 022 qmult 00530 1 1 3 easy memory: Dirac comb 5. A periodic potential that consists a periodic array of equally strong Dirac delta function potential spikes separated by flat potential reigons is called a comb. a) Bloch

b) Compton

c) Dirac

d) Fermi

e) Pauli

Full-Answer Problems 022 qfull 00100 2 5 0 moderate thinking: free electron solid Extra keywords: (Ha-324:2.4) 1. Let us consider a free electron gas model of a solid in 1, 2, and 3 dimensions simultaneously. Use periodic boundary conditions (AKA Born-Von-Karman boundary conditions) and assume cubical shape in all three cases: a 1-dimensional cube is a line segment, a 2-dimensional cube is a square, and a 3-dimensional cube is a cube. Let L be the length of a side of the cube and Lℓ be the cube volume where ℓ is the number of dimensions. a) Solve the time-independent Schr¨odinger equation for the single-particle stationary states for all three cases. Normalize the solutions solutions and give their quantization rules for wavenumbers and energy. It is a given that the multi-particle time-independent Schr¨odinger equation separates into identical single-particle time-independent Schr¨odinger equations. But it is NOT a given that the single-particle time-independent Schr¨odinger equation separates into 3 identical 1-dimensional time-independent Schr¨odinger equations. b) What is the volume Vk in k-space of cubes that are centered on the allowed spatial state wave vectors and that are contiguous with each other: i.e., what is the k-space volume per state? What is the average density of spatial states in k-space ρk ? c) We now make the continuum approximation which is valid for samples that are macroscopic in all available dimensions. This means that we treat the average density of spatial states in k-space as if it were an uniform density. Find the expression for the differential number of states dN in a spherical shell in k-space. The shell radius is k and its thickness is dk. Include the spin degeneracy by a factor g which equals 2 for spin 1/2 electrons. But leave g unevaluated. By leaving g unevaluated, one can track how the spin degeneracy affects dN and expressions derived from dN . d) The Pauli exclusion principle for fermions requires that each single particle spatial-spin state have only one fermion at most. This statement must be qualified. What it really means is that the product wave function of single particle states can have each distinct single-particle state included once only. If there are M fermions, the overall symmetrized wave function contains M ! versions of the product wave function with the individual particle coordinate labels in all possible permuations. But we don’t have to worry about product wave functions or symmetrized wave functions explicitly in the free electron gas model of solids. We simply make use of the Pauli exclusion principle to say that the single-particle states can only be used once in calculating results or to put this in common, but somewhat misleading, jargon only one electron can occupy at single particle state at most.

138 Chapt. 22 Solids Now in the ground state of (which is the absolute zero temperature state) in our free electron gas model, the electrons occupy the lowest energy single-particle states consistent with the Pauli exclusion principle. This means in k-space, the electrons occupy a sphere of radius kF where F where stands for Fermi: since F stands for a name, not a variable it ought to be in Roman, not Italic, font, but convention seems to dictate Italic font (e.g., Gr-221, CT-1435). The radius kF is called the Fermi wavenumber. Using the results of the part (c) answer solve for kF for an electron density (in space space) of ρe = M/Lℓ . e) Now solve for the Fermi energy EF for the 3 cases. f) What is ρE : i.e., the density of states per unit volume per unit energy in the continuum of states approximation. Write a general formula that is valid for all three dimension cases. HINT: One requires the same number of states between any corresponding limits: i.e., dN = ρk shell dk = ρE dE , where ρk shell is the density of spatial-spin states in the differential k space shell dk. The general expression for ρk shell must have turned up in the part (c) answer without so labeling it. g) Now solve for the total energy Etotal of the ground state for M electrons, Eave the average energy for the M electrons, and Eave /EF . Don’t bother to expand EF using the expressions from the part (e) answer. The formulae for this answer are long enough as it is. 022 qfull 00110 1 3 0 easy math: free electron gas formulae and fiducial formulae 2. For a free electron gas (in 3 dimensions) at abolute zero temperature, the Fermi energy is given by 2/3 2   − 2 h 2 , (3π )ρe EF = 2m g where − h is Planck constant divided by 2π, m is the electron mass, g = 2 is the spin 1/2 particle degeneracy, and M ρe = V is the free electron density with M being the number of electrons and V being the sample volume. The average energy per electron Eave is given by Eave =

3 EF . 5

For this question, you will need the following constants, e = 1.602176487(40) × 10−19 C , m = 9.10938215(45) × 10−31 kg ,

mc2 = 510998.910 eV , kB = 1.3806504(24) × 10−23 J/K , − h = 1.054571628(53) × 10−34 J, s ,

mamu = 1.660538782(83) × 10−27 kg .

a) Free electron density can be expressed in terms of ordinary density (AKA mass density) ρ by ρ X Xi Zi , ρe = mamu i Ai

Chapt. 22 Solids 139 where the sum is over all atoms in the sample, Xi is the mass fraction of atom i, Zi is the number of free electrons per atom for atom i, and Ai is the atomic weight of atom i. Convince yourself that this formula makes sense. Actually the formula can be simplified by introducing the mean mass per electron µe defined by X Xi Zi 1 = . µe Ai i

Take 1000 kg/m3 as a fiducial value for ρ (which is just like writing density in grams per cubic centimeter). Take 50 as a fiducial value for µe : this is like an element in the atomic weight vicinity of iron with one valence electron delocalized. Now write ρe in terms of fiducial values: i.e., find the coefficient in the formula expression ρe = coefficient ×

ρ 50 , 3 1000 kg/m µe

The coefficient is a fiducial electron density. b) Now find the formula for EF in terms of fiducial values both for joules and electronvolts: i.e., find the coefficient in the formula 2/3   ρ 50 2 EF = coefficient × . g 1000 kg/m3 µe The coefficient is a fiducial Fermi energy. Are solids in ordinary human environments relativistic? c) Now find the formula for the Fermi temperature TF = EF /kB in terms of fiducial values both for joules and electronvolts. The coefficient is a fiducial Fermi temperature. Are solids are ordinary human environments hot or cold in comparison to the Fermi temperature? p d) Now find the formula for the Fermi velocity vF = 2EF /m (which is the non-relativistic formula) in terms of fiducial values. The coefficient is a fiducial Fermi velocity. e) For zero temperature, P =−

∂E ∂V

where E is the total energy of the sample and V is the sample volume. A free electron gas exhibits a pressure derivable from this classical expression. The pressure is called the degeneracy pressure and it’s equation of state is quite unlike that for an ideal gas. Derive the zero-temperature free-electron-gas pressure formula as a function of ρe . Then find the pressure formula in terms of fiducial values. f) The bulk modulus of a material is a measure of its incompressibility or stiffness. The definition is dP dP =ρ B = −V dV dρ where temperature needs to be specified in general and were we have used dV = d which can also be written

  1 1 V = − 2 dρ = − dρ ρ ρ ρ dρ dV =− . V ρ

140 Chapt. 22 Solids One can see that bulk modulus is a measure stiffness by rewriting to dP dρ = . ρ B The bulk modulus is the coefficient needed to convert a differential change in pressure in the corresponding RELATIVE differential change in density. The bigger the bulk modulus, the stiffer the substance. Find the bulk modulus formula for the zero-tempature free electron gas. Then find the bulk modulus formula in terms of fiducial values. 022 qfull 00200 2 5 0 moderate thinking: computing Fermi energies Extra keywords: (Ha-324:2.3) free electron metals 3. The metals Na, Mg, and Al have, respectively 1, 2, and 3 free electrons per atom and volumes per atom 39.3 ˚ A3 , 23.0 ˚ A3 , and 16.6 ˚ A3. At zero temperature what is the Fermi energy to which the free electron states are filled? 022 qfull 00500 1 3 0 easy math: energy band structure function Extra keywords: Bloch states 4. The energy band structure of the 1-dimensional Dirac comb solid is determined by the equation cos(Ka) = cos(z) +

β sin(z) , z

where a is the cell size, K is the Bloch wavenumber, z = ka ≥ 0 (where k is the wavenumber for a single cell), and mαa β= 2 − h (where m is electron mass and α is the strength of the Dirac delta function potentials that make up the Dirac comb). The Bloch wavenumbers are quantized according to the rule Ka =

2π m, N

where m = 0, 1, 2, 3, . . . , N − 1 (where N is the number of cells). Actually, Ka is not determined to within an additive constant of 2π. Therefore one could also specify the range for m as m=−

N N N N , − + 1, . . . , − 1, 2 2 2 2

since for N large one can with negligible error approximate N as even no matter what it actually is and add on an extra unphysical state. This second specification for m is more symmetrical and, as it turns out, makes k vary monotonically with K within a band. For definiteness in this problem, we assume β ≥ 0. The only possible solution z values are those that make confine f (z) = cos(z) +

β sin(z) z

to the range [−1, 1]. For a Ka value, one solves cos(Ka) = f (z) for ka and then k. Thus, the k values have a non-trivial quantization rule. The energy of the k state is then determined by 2 − h k2 . E= 2m

Chapt. 22 Solids 141 The energies have a non-trivial quantization rule too. A key point is the function f (z) can oscillate above 1 and below −1 and in those regions there are no solutions for k, and therefore no allowed states and no solutions for E. Those energy regions are the famous energy gaps. The allowed regions are the energy bands. Since N is an enormous number for macroscopic samples, the energy bands are very dense in number of states and one gets states virtually to the z points where |f (z)| = 1. The z values that yield |f (z)| = 1 are the band limits. Each band has a start and end limit. There are corresponding energy start and end limits. a) Determine the formula end limit of a general ℓth energy band. Note ℓ = 1, 2, 3, . . . Show that the formula is the end limit formula and never gives start limits. HINT: Finding the formula is easy. Showing that it is the end limit requires taking the derivative of f (z). b) A key point to proof is that there is only one limit point between two consecutive end limit points given by the formula in the part (a). This single limit point is a start limit point. In a test situation just assume there is a single start limit point between the end limit points and go onto the rest of the problem. In a non-test situation, do the proof. HINT: Prove there is a single stationary point between the consecutive end limit points. The function f (z) value at the stationary point is must be outside of the range [−1, 1] by the part (a) answer. Also from the part (a) answer, the function at the stationary point must be greater than 1 if the next end limit point f (z) = −1 and the function at the stationary point must be less than 1 if the next end limit point is f (z) = 1 Thus from the stationary point to the next end limit point the function is is monotonic and crosses into the allowed range once only. That is crossing is the start limit point itself. The stationary point itself can’t be found analytically in general, but its uniqueness can be proven by arranging the equation for stationary point into a form where tan(z)/z equals something. c) Find an approximate formula for the start limit for the first band in the limit that β is very large. Show that this formula also gives the start limit for β = 0, and thus constitutes an interpolation formula: a formula that gives correct limiting behavior and interpolates smoothly in between those limits. HINT: Set the function f (z) equal to 1 and expand the trigonometric functions in f (z) about π to 1st order in small z − π. Why are the expansions the good thing to do and why is it good not to expandi the 1/z factor to 1st order in small z − π even though it would be consistent with the other expansions? d) Find an approximate formula for the start limit ∆z = z − zℓ−1 = z − (ℓ − 1)π for the general ℓth band, except that ℓ ≥ 1. HINT: Set f (z) = (−1)ℓ−1 and expand cos(z) to second order in small ∆z about zℓ−1 . and sin(z) to 3rd order in small ∆z about zℓ−1 = (ℓ−1)π Do NOT approximate 1/(zℓ−1 + ∆z. Write β = yzℓ−1 /2 as a simplification. What approximate size limit must be put on ∆z for our approximations to be valid? What is ∆z in the limit that zℓ−1 becomes very large. e) Take the interpolation formula from the part (c) answer the ∆z formula for very large zℓ−1 from the part (d) answer and invent an interpolation formula for ∆z valid for all ℓ. We mean it is valid for ℓ in that it gives the right limiting behavior for β for the ℓ = 1 and the right limiting behaviro for ℓ becoming very large. Since it has those right limiting behaviors, it may well interpolate to some accuracy everywhere. HINT: There is no absolutely right answer, but a fairly obvious interpolation formula leaped to yours truly’s eye. 022 qfull 00510 1 3 0 easy math: energy band structure function 2 5. This a super-easy problem if you can understand the lengthy setup. For the 1-dimensional Dirac comb potential, the band structure is determined by the equation cos(Z) = f (z) ,

142 Chapt. 22 Solids where

β sin(z) . z The Z is determined by the Bloch wavenumber quantization rule f (z) = cos(z) +

Z = Ka =

2π m N

where m = 0, 1, 2, . . . , (N − 1)

where N is the number of number of cells and a is the cell length. For a macroscopic system N is huge, and so Z can be close to any real number in the range [0, 2π]. The z = ka, where k is the ordinary wavenumber from which the energy of a single-particle state can be determined from 2 − h k2 E= . 2m If we knew a z value, determining the corresponding Z value would be a cinch. But we know the Z values and need to determine the z values. There is no analytic solution for general z. We do know that the only solutions are in bands set by the fact that cos(Z) can only be in the range [−1, 1]. From earlier work, we know that zℓ,start = (ℓ − 1)π + ∆zℓ = (ℓ − 1)π +

2βπ 2β + 4 + π 2 (ℓ − 1)

is a good approximation for the start limit of band ℓ and zℓ,end = ℓπ is the exact result for the end limit of a band ℓ. The band numbers run form ℓ = 1 to ℓ = ∞. The cos(Z) value sweeps from 1 to −1 as z increases for odd bands and from −1 to 1 for even bands. Let’s define a parameter g that runs from 0 to 1 and parameterize Z in terms of g thusly Z=



gπ (g + 1)π

for ℓ odd; for ℓ even.

Using g (which we can easily find for any Z value and any ℓ) and the formulae for zℓ,start and zℓ,end devise an approximate z formula that is LINEAR in g. HINT: This is super-easy, but you should test that g = 0 and g = 1 give the correct band limits: z(ℓ, g = 0) = zℓ,start and z(ℓ, g = 1) = zℓ,end.

Chapt. 23 Interaction of Radiation and Matter

Multiple-Choice Problems 023 qmult 00100 1 1 4 easy memory: Maxwell’s equations 1. Classical electrodynamics is summarized in: a) b) c) d) e)

Stirling’s formula. Heitler’s five equations. Montefeltro’s three laws. Maxwell’s four equations. Euclid’s axioms.

023 qmult 00200 1 4 5 easy deducto-memory: EM potentials 2. “Let’s play Jeopardy! For $100, the answer is: ~ ~ = −∇Φ − 1 ∂ A E c ∂t a) b) c) d) e)

and

~ =∇×A ~ .” B

What are Maxwell’s equations, Alex? What are two arbitrary equations, Alex? What are the gauge transformation equations, Alex? What are the Noman equations, Alex? What are the equations for deriving the electromagnetic fields from the electromagnetic potentials, Alex?

023 qmult 00300 1 1 3 easy memory: transverse gauge 3. The condition imposed for the transverse gauge (or radiation or Coulomb gauge) is: a) b) c) d) e)

~ = 0. ∇·B ~ + (1/c)∂Φ/∂t = 0. ∇·A ~ = 0. ∇·A ~′ = A ~ + ∇Λ. A Φ′ = Φ − (1/c)∂Λ/∂t.

023 qmult 00400 1 4 2 easy deducto-memory: EM energy density Extra keywords: Reference Ja-236 4. “Let’s play Jeopardy! For $100, the answer is: E= a) b) c) d)

What What What What

E2 + B2 .” 8π

Laplace’s equation, Alex? the energy density of the electromagnetic field in SI units, Alex? the energy density of the electromagnetic field in Gaussian units, Alex? Poynting’s vector, Alex? 143

144 Chapt. 23 Interaction of Radiation and Matter e) What is an example of using “E” for two quantities in one equation, Alex? 023 qmult 00500 1 1 2 easy memory: wave equation solution 5. The simple wave equation 1 ∂ 2~g ∇2~g = c ∂t2 has a very general solution of the form: a) b) c) d) e)

~g = 0. ~g (~k · ~r − ωt), where ~g is any vector function and ω = |~k|c. ~g (~k × ~r − ωt), where ~g is any vector function and ω = |~k|c. ~g (ωt). ~g (~k · ~r).

023 qmult 00600 1 1 2 easy memory: EM perpendicular vectors 6. For self-propagating electromagnetic radiation the electric field, magnetic field, and propagation directions are: a) b) c) d) e)

collinear. mutually perpendicular. self-perpendicular. arbitrary. random.

023 qmult 00700 1 4 4 easy deducto-memory: box quantization 7. “Let’s play Jeopardy! For $100, the answer is: A standard trick in quantum mechanics for replacing basically isotropic, homogeneous systems with complex boundary conditions by a simpler, tractable, periodic-boundary-condition system which—according to the faith—must have the same bulk properties.” a) b) c) d) e)

What What What What What

is is is is is

nothing I’ve ever heard of, Alex? the tethered function method, Alex? black magic, Alex? box quantization, Alex? the way to get a completely wrong answer, Alex?

023 qmult 00800 1 4 3 easy deducto-memory: interaction Hamiltonian 8. “Let’s play Jeopardy! For $100, the answer is: H=

e ~ 2 1  p~ − A + eΦ + V .” 2m c

a) What is Schr¨odinger’s equation, Alex? b) What is the Hamiltonian for the simple harmonic oscillator, Alex? c) What is the particle Hamiltonian including the interaction with the electromagnetic field, Alex? d) What is the interaction Hamiltonian of the electromagnetic field, Alex? e) What is any old Hamiltonian, Alex? 023 qmult 01200 1 1 4 easy memory: cross section 9. What one often wants from a radiation interaction calculation is a coefficient or the like that can be employed in macroscopic radiative transfer: e.g., a) a postulate.

Chapt. 23 Interaction of Radiation and Matter 145 b) c) d) e)

a a a a

postulant. golden section. cross section. crossbow.

023 qmult 01300 1 1 1 easy memory: quantizing radiation 10. To quantize the electromagnetic radiation field we assert that the energy of radiation mode cannot take on a continuum of values, but must come in: a) b) c) d) e)

integral amounts of a basic unit called a quantum of electromagnetic radiation or a photon. amounts related by the golden section. photoids. integral numbers of ergs. integral numbers of eVs.

023 qmult 01400 1 1 3 easy memory: creation and annihilation operators 11. The electromagnetic field operator is defined in terms of: a) b) c) d) e)

Bell and Sprint operators. left and right operators. creation and annihilation operators. procreation and inhibition operators. genesis and revelations operators.

023 qmult 01500 1 1 2 easy memory: zero point energy 12. In the formalism of quantized electromagnetic radiation, the fact that the creation and annihilation operators don’t commute leads to: a) b) c) d) e)

nothing in particular. the existence of zero-point energy. energy less than the zero-point energy. living close to the office. living in the office.

023 qmult 01600 1 4 4 easy deducto-memory: Einstein stimulated em. 13. “Let’s play Jeopardy! For $100, the answer is: An effect discovered by Einstein by means of a thermodynamic detailed balance argument.” a) b) c) d) e)

What What What What What

is is is is is

spontaneous emission, Alex? special relativity, Alex? the photoelectric effect, Alex? stimulated emission, Alex? spontaneous omission, Alex?

023 qmult 01700 1 4 5 easy deducto-memory: free particles don’t radiate 14. “Let’s play Jeopardy! For $100, the answer is: A particle that cannot radiate by a Fermi golden rule process in the non-relativistic limit and perhaps not at all—but one never knows what special exotic cases may be said to allow it to radiate.” a) b) c) d) e)

What What What What What

is is is is is

a classical particle, Alex? a neutral pion, Alex? a quark, Alex? an unbound particle subject to potentials, Alex? a free particle, Alex?

146 Chapt. 23 Interaction of Radiation and Matter 023 qmult 01800 1 1 1 easy memory: conservation of momentum 15. Since it seems that no quantum mechanical matter particle can actually be in a pure momentum eigenstate, conservation of momentum in radiative emission or absorption in the instructor’s view a) b) c) d) e)

cannot be strict in a classical sense. can be strict in a classical sense. both can and cannot be strict in a classical sense. can neither be nor not be strict in a classical sense. is moot and categorically valid in a classical sense.

023 qmult 01900 1 1 2 easy memory: electric dipole transitions 16. Typically, strong atomic and molecular transitions are a) b) c) d) e)

electric quadrupole transitions. electric dipole transitions. magnetic dipole transitions. electric monopole transitions. magnetic metropole transitions.

023 qmult 02000 1 4 4 easy deducto-memory: selection rules 17. “Let’s play Jeopardy! For $100, the answer is: The selection rules for electric dipole transitions (i.e., allowed transitions) between energy eigenstates whose angular parts are spherical harmonics.” a) b) c) d) e)

What What What What What

are are are are are

Hund’s rules, Alex? ∆m = ±2 and ∆ℓ = 0, Alex? ∆m = ±1 and ∆ℓ = 0 or ±1, Alex? ∆m = 0 or ±1 and ∆ℓ = ±1, Alex? taking the fresher and firmer ones, Alex?

Full-Answer Problems 023 qfull 00300 2 5 0 moderate thinking: classical EM scattering Extra keywords: reference Mi-83 1. Say we had a classical simple harmonic oscillator (SHO) consisting of a particle with mass m and charge e and a restoring force mω02 where ω0 is the simple harmonic oscillator frequency. This SHO is subject to driving force caused by traveling electromagnetic field (i.e., light): ~ 0 eiωt , F~drive = eE ~ 0 is the amplitude, ω is the driving frequency, and we have used the complex exponential where E form for mathematical convenience: the real part of this force is the real force. The magnetic force can be neglected for non-relativistic velocities. The Lorentz force is   ~ ~ + ~v × B F~ = e E c ~ and B ~ are comparable in size for electromagnetic radiation, and so the magnetic (Ja-238) and E force is of order v/c smaller than the electric force. (See also MEL-130.) An oscillating charge is an accelerating charge and will radiate electromagnetic radiation. The power radiated classically is 2e2 a2 , P = 3c3

Chapt. 23 Interaction of Radiation and Matter 147 where ~a is the charge acceleration. This radiation causes an effective damping force given approximately by F~damp = −mγ~v , where

2e2 ω02 . 3mc3 The full classical equation of motion of the particle is γ=

~ 0 eiωt − mγ~v . m~a = −mω02~r + eE a) Solve the equation of motion for ~r and ~a. HINTS: The old trial solution approach works. Don’t forget to take the real parts although no need to work out the real part explicitly: i.e., Re[solution] is good enough for the moment. b) Now solve for the time average of the power radiated by the particle. HINT: You will need the explicit real acceleration now. c) The average power radiated must equal the average power absorbed. Let’s say that the particle is in radiation flux from a single direction with specific intensity I0 =

cE02 8π

(Mi-9), where time averaging is assumed como usual. The power absorbed from this flux is σ(ω)I0 , where σ(ω) is the cross section for energy removed. Solve for σ(ω) and then find show that it can be approximated by a Lorentzian function of ω with a coefficient πe2 /(mc). HINT: It is convenient to absorb some of the annoying constants into another factor of γ. d) Now rewrite the cross section as a function of ν = ω/(2π) (i.e., the ordinary frequency) and then integrate over ν to get the frequency integrated cross section σν int of the system. What is the remarkable thing about σν int ? Think about how it relates to the system from which we derived it. Evaluate this frequency integrated cross section for an electron. HINT: The following constants might be useful 1 e2 α= − = , 137.036 hc

− h = 1.05457 × 10−27 erg s ,

and

me = 9.10939 × 10−28 g .

023 qfull 00500 2 5 0 moderate thinking: gauge invariance Extra keywords: (Ba-299:1) 2. The gauge transformations of the electromagnetic potentials are:

and

~ r , t)′ = A(~ ~ r , t) + ∇χ(~r, t) A(~ φ(~r, t)′ = φ(~r, t) −

1 ∂χ(~r, t) c ∂t

~ r , t) is the vector potential, φ(~r, t) is the scalar potential, the primed (Ja-220), where A(~ quantities are the transformed versions, and χ(~r, t) is the gauge. Show that the solution to the time-dependent Schr¨odinger equation for n particles of charge e and mass m transforms so # " n X ie χ(~ri , t) Ψ(~r1 , . . . , rn , t) Ψ(~r1 , . . . , ~rn , t)′ = exp − h c i=1

148 Chapt. 23 Interaction of Radiation and Matter under a gauge transformation of the potentials treated as classical fields: i.e., if Ψ′ satisfies a primed version of the Schr¨odinger equation, then Ψ satisfies an unprimed version. HINT: It suffices to consider one particle in one dimension. I know of no simplifying tricks. You just have to grind out the 15 odd terms very carefully. 023 qfull 00700 2 5 0 moderate thinking: QM CA operators Extra keywords: (Ba-299:3) 3. The photon creation and destruction operators of the electromagnetic field photons (the CA operators for short) are defined by, respectively s 2π− h c2 q N~k~λ + 1| . . . , N~k~λ + 1, . . .i A~†k~λ | . . . , N~k~λ , . . .i = ω and s 2π− h c2 q A~k~λ | . . . , N~k~λ , . . .i = N~k~λ | . . . , N~k~λ − 1, . . .i , ω where N~k~λ is the number of photons in the box quantization mode specified by wavenumber ~k and polarization vector ~λ. All the physically unique states created by these operators are assumed to be orthogonal. From this assumption it follows that creation and annihilation operators are Hermitian conjugates as their labeling indicates: we’ll leave that proof sine die. The annihilation operator acting on the empty mode (i.e., one with N~k~λ = 0) gives a zero or null state (or vector). Any operator acting on a zero state gives a zero state, of course. Thus, non-zero commutator identities can’t be proven by acting on zero states and it is understood that zero states are never thought of in proving commutator identities.

and

Dimensionless forms of these operators are, respectively, p a†i | . . . , Ni , . . .i = Ni + 1| . . . , Ni + 1, . . .i ai | . . . , Ni , . . .i =

p Ni | . . . , Ni − 1, . . .i ,

where index i subsumes both the ~k and ~λ indices. a) Prove the commutator identities [ai , aj ] = 0 ,

[a†i , a†j ] = 0

and

[ai , a†j ] = δij .

Remember the case of i = j for the first two identities. What are the dimensioned forms of these commutator identities? b) Recall that classically the total energy in the radiation field is  Z  2 E + B2 d~r , E= 8π where calligraphic E is total energy, E is the electric field, B is the magnetic field (or magnetic induction if you prefer), and the integration is implicitly over all the volume of the system. The quantum mechanical, Heisenberg representation field operator for the radiation field is " # i(~ k·~ r −ωt) X e ~ op = A~k~λ~λ √ A + H.C. , V ~~ kλ

where we have assumed box quantization with periodic boundary conditions, V is the volume of the box, and “H.C.” stands for Hermitian conjugate. What is the time-averaged

Chapt. 23 Interaction of Radiation and Matter 149 Heisenberg representation of the Hamiltonian of the electromagnetic field Hem in terms of the operator A~†k~λ A~k~λ ? c) What is Hem in terms of dimensionless EM operators? What is the zero point energy of a radiation field: i.e., the energy expectation value for the state |0, 0, 0, . . . , 0i? d) Convert the summation for the zero point energy into an integral using the continuum approximation for the box quantization. What is the energy density E0 /V of the vacuum up to the mode with photon energy ǫ? (Note that ǫ is not the energy in the mode, it is the energy of photons in the mode: each photon has ǫ = − h ω.) What unfortunate thing happens if you set ǫ = ∞? 023 qfull 00800 2 5 0 moderate thinking: commutation field operators Extra keywords: (Ba-299:4) 4. The quantum mechanical, Heisenberg representation field operator for the radiation field is # " ~ X ei(k·~r−ωt) ~ ~ + H.C. , A~k~λ λ √ Aop (~r, t) = V ~~ kλ

where we have assumed box quantization with periodic boundary conditions, V is the volume of the box, and “H.C.” stands for Hermitian conjugate. ~ op (~r, t) and B ~ op (~r, t). a) Determine expressions for E ~ op (~r, t), B ~ op (~r, t)]. What can you say about the simultaneous knowledge that b) Determine [E ~ and B? ~ HINTS: The following commutator relations will help one can have about E simplify: [A~k~λ , A~k′ ~λ′ ] = 0 ,

[A~† ~ , A~† ′ ~ ′ ] = 0 , kλ

k λ

and

[A~k~λ , A~† ′ ~ ′ ] = δ~k~k′ δ~λ~λ′ k λ

2π− h c2 , ω

where ω = kc, of course. You will also have to deal with the outer product of two vectors which is a tensor: e.g., ~a ⊗ ~b. The outer product is commutative: e.g., ~a ⊗ ~b = ~b ⊗ ~a (e.g., ABS-25). 023 qfull 01500 2 5 0 moderate thinking: free particles non-radiate Extra keywords: Reference Ba-280 5. Say you have a free particle (i.e., a free matter particle) in a momentum eigenstate with eigen momentum − h~ qn . Remember a free particle in the conventional sense of quantum mechanics means one not affected by any potentials at all. Using Fermi’s golden rule you can try to calculate the particle’s spontaneous emission when making a transition to another momentum eigen state with eigen momentum ~q0 (Ba-279). You find a momentum conservation rule is imposed: −~ h ~qn − − h ~q0 , hk = − where −~ h k is the momentum of the emitted photon. But since you used the golden rule you also imposed energy conservation: 2 2 − − h qn2 h q02 − h ck = − , 2m 2m where m is the particle’s mass. The conclusion that is implied by these two relations is that free particles can’t radiate at least not through the golden rule process used in the calculations. Show how the conclusion is reached. HINT: The two relations actually can be mutually satisfied in a mathematical sense: think about the assumptions implicit in them.

150 Chapt. 23 Interaction of Radiation and Matter 023 qfull 02000 3 5 0 moderate thinking: hydrogen spontaneous emission Extra keywords: (Ba-300:7) 6. Consider a hydrogen atom with the nucleus fixed in space. a) What are all the states of the two lowest principal quantum numbers in spectroscopic notation: e.g., 2s(m = 0) for n=2, ℓ=0 (which is what the s symbol means), and m = 0. Don’t distinguish states by spin state. b) Between which states are there (electric dipole) allowed transitions transitions. c) The spontaneous emission power per solid angle in polarization ~λ in the electric dipole approximation is dPλ ω 4 e2 ~ |d0n · ~λ∗ |2 , = dΩ 2πc3 where

~ d~0n = h0|R|ni

is the dipole moment matrix element, |ni and |0i are, respectively, the initial and final states, and X ~ = ri R i

is the position operator for all the particles in the system (Ba-282). Find the expression for d~0n between the 1s(m = 0) and 2p(m = 0) levels of hydrogen. 023 qfull 02500 2 5 0 moderate thinking: Einstein A coefficient Extra keywords: See Ba-282, Gr-311–313 7. In the electric dipole approximation, the power in polarization λ per unit solid angle per particle (or system of particles) of a spontaneous transition process is dPλ ω 4 e2 ~ |d0n · ~λ∗ |2 , = dΩ 2πc3 where e is the charge on the particle (or particles), ~ d~0n = h0|R|ni is the off-diagonal element of the dipole moment operator, and n and 0 label the initial and final states, respectively (Ba-282). In general, ~ = R

X

~ri .

i

where the sum is over the position operators of the particles in the system. a) Consider an ordinary 3-d Cartesian set of axes and radiation emitted along the z-axis. Consider the polarization vectors to be unit vectors in the x and y directions. The dipole matrix element points in an arbitrary direction (θ, φ) in spherical polar coordinates. What is the expression for power SUMMED over the two polarization directions in terms of the angle coordinates of the dipole matrix element? HINTS: A diagram probably helps and knowing how to express d~0n along Cartesian coordinates, but in spherical polar coordinates. b) Using the expression from the part (a) answer, find the total power radiated by integrating over all solid angle. HINT: Nothing forbids you from mentally transposing the z-axis from the general direction to a convenient direction.

Chapt. 23 Interaction of Radiation and Matter 151 c) What is the total photon number rate emitted from the transition (i.e., the Einstein A coefficient at least as Gr-312 defines it)? d) If at time zero you have set of N0 atoms in an excited state with only one downward transition available with Einstein A coefficient A and spontaneous emission as the only process, what is the population N at any time t > 0?

Chapt. 24 Second Quantization

Multiple-Choice Problems 024 qmult 00100 1 1 2 easy memory: 2nd quantization 1. The formalism for quantizing fields is called: a) b) c) d) e)

first quantization. second quantization. third quantization. fourth quantization. many quantization.

024 qmult 00200 1 4 3 easy deducto-memory: CA anti/commutation 2. “Let’s play Jeopardy! For $100, the answer is: These operators for a single mode have commutation relation [a, a† ] = 1 for bosons and anticommutation relation {a, a† } = 1 for fermions.” a) b) c) d) e)

What What What What What

are are are are are

Hermitian operators, Alex? anti-Hermitian operators, Alex? creation and annihilation operators, Alex? penultimate and antepenultimate operators, Alex? genesis and revelations operators, Alex?

024 qmult 00300 1 1 1 easy memory: boson CA operator effects 3. For bosons the operators a and a† acting on a state |ni give, respectively: √ √ a) √n|n − 1i (or the null vector if n = 0) and n + √ 1|ni. b) √n − 1|n − 1i (or the null vector if n = 0) and√ n|ni. c) n| − n + 1i (or the null vector if n = 0) and −n| − n + 1i. d) the null vector and the infinite vector. e) a non-vector and a non-non-vector. 024 qmult 00400 1 1 4 easy memory: fermion number operator 4. What are the eigenvalues of the fermion number operator obtained from the matrix representation this operator: i.e., from N=

a) b) c) d) e)



0 0 0 1

±1 Both are zero. 1 and 2. 0 and 1. ±2. 152



?

Chapt. 24 Second Quantization 153 024 qmult 00500 1 1 1 easy memory: number operators commute 5. The number operators for different modes for the boson and fermion cases: a) b) c) d) e)

commute. anticommute commute and anticommute, respectively. anticommute and commute, respectively. promote and demote, respectively.

024 qmult 00600 1 1 2 easy memory: CA operators for symmetrization 6. Symmetrized states in second quantization formalism are easily constructed using a) b) c) d) e)

annihilation operators. creation operators. more annihilation operators than creation operators. steady-state operators. degenerate operators.

024 qmult 00700 1 4 5 easy deducto-memory: field operators 7. “Let’s play Jeopardy! For $100, the answer is: These operators are constructed from the creation and annihilation operators in second quantization formalism.” a) b) c) d) e)

What What What What What

are are are are are

the the the the the

copse operators, Alex? glade operators, Alex? meadow operators, Alex? field marshal operators, Alex? field operators, Alex?

024 qmult 01000 1 4 4 easy deducto-memory: 2nd QM density operator Extra keywords: Reference Ba-422 8. “Let’s play Jeopardy! For $100, the answer is: ρ(~r ) = Ψs (~r )† Ψs (~r ) , where Ψs (~r )† and Ψs (~r ) are the field creation and annihilation operators at a point ~r with spin coordinate s.” a) b) c) d) e)

What What What What What

is is is is is

a non-Hermitian operator, Alex? the density expectation value, Alex? the first quantization density operaty, Alex? the second quantization particle density operator, Alex? it’s Greek to me, Alex?

024 qmult 01010 1 1 1 easy memory: density operator expectation value Extra keywords: See Ba-422 9. There is a second quantization operator ρ(~r ) = Ψs (~r)† Ψs (~r ) , where Ψs (~r )† and Ψs (~r ) are the field creation and annihilation operators at a point ~r with spin coordinate s. Its expectation value is: a) the mean density of particles per unit volume. b) the normalization constant of the state |Φi IT is applied to. c) the pair correlation function.

154 Chapt. 24 Second Quantization d) the pair anti-correlation function. e) the one-particle density matrix. 024 qmult 01100 1 1 3 easy memory: zeros of one-particle density matrix Extra keywords: See Ba-426 10. The one-particle density matrix for non-interacting, spin 1/2 fermions in the ground state of a box quantization system is Gs (~r − ~r ′ ) =

3n sin(x) − x cos(x) , 2 x3

where n is the expectation particle density of the ground state and x = pf |~r − ~r ′ | (i.e., Fermi momentum times displacement vector). The zeros of this function are approximately given by: a) b) c) d) e)

x = n, where n = 0, 1, 2, . . . x = nπ, where n = 1, 2, 3, . . . x = (1/2 + n)π, where n = 1, 2, 3, . . . x = (1/2 + n)π, where n = 0, 1, 2, . . . x = nπ, where n = 0, 1, 2, . . .

024 qmult 01300 1 1 2 easy memory: pair correlation function Extra keywords: See Ba-429 11. For a box quantization system of non-interacting, spin 1/2 fermions in the ground state we have the useful function ( 1 for s 6= s′ ; ′ 9 gss′ (~r − ~r ) = 1 − 6 [sin(x) − x cos(x)]2 for s = s′ . x where and x = pf |~r − ~r ′ | (i.e., Fermi momentum times displacement vector). This function is the: a) b) c) d) e)

pair anti-correlation function. pair correlation function. pair annihilation function. pair creation function. one-particle density matrix.

024 qmult 01700 1 4 3 easy deducto-memory: exchange effect 12. “Let’s play Jeopardy! For $100, the answer is: This effect generally decreases the absolute value of the potential energy of an interaction between fermions of the same spin coordinate—except in the unusual case that the interaction increases with INCREASING distance between the fermions.” a) b) c) d) e)

What What What What What

is is is is is

expunge effect, Alex? the interchange effect, Alex? the exchange effect, Alex? the interstate effect, Alex? the exchange rate effect, Alex?

024 qmult 02000 1 1 2 easy memory: Feynman diagrams Extra keywords: See Ha-211 13. As a visualization of the terms in an interaction perturbation series in second quantization formalism, one can use: a) Feynman landscapes.

Chapt. 24 Second Quantization 155 b) c) d) e)

Feynman Feynman Feynman Feynman

diagrams. water colors. lithographs. doodles.

Full-Answer Problems 024 qfull 00100 2 5 0 moderate thinking: boson CA proofs 1. Let’s do proofs with the boson creation and annihilation (CA) operators a and a† for a single mode. Recall [a, a† ] = 1 is their fundamental commutation relation. a) The number operator is defined by N = a† a. Show that N is Hermitian. Assume N is an observable (i.e., a Hermitian operator with a complete set of eigenstates for whatever space we are dealing with) and let its eigen-equation be N |ni = n|ni . We also assume there is no degeneracy. Show that n ≥ 0 and that a|0i = 0 (i.e., a|0i is the null vector). HINT: Make use of the rule—which I think is valid—at least lots of sources use it or so I seem to recall—that a legitimate operator acting on a vector always yields a vector. b) Find explicit expressions for [N, a] and [N, a† ]. c) Show that a† |ni and a|ni are eigenstates of N and find explicit expressions for them. d) Show that n must be an integer. e) Find a general expression for |ni in terms of the vacuum state |0i. Make sure that |ni is properly normalized. f) Since the |ni are non-degenerate states of the number operator N = a† a (an observable), they are guaranteed to be orthogonal. But for the sake of paranoia vis-`a-vis the universe, show explicitly that hm|ni = δmn , making use of the part (e) answer and assuming the vacuum state is properly normalized. HINT: This is easy after you’ve seen the trick. g) Prove [a, (a† )n ] = n(a† )n−1 for n ≥ 1. 024 qfull 00200 2 5 0 moderate thinking: 2 mode fermion CA operators Extra keywords: (Ba-439:1) 2. Let us consider the two-mode fermion case. a) Construct explicit 4 × 4 matrix representations of the creation and annihilation (CA) operators a0 , a†0 , a1 , and a†2 . HINT: See Ba-414–415. b) Construct explicit 4 × 4 matrix representations of the number operators N0 , N1 , and N = N0 + N1 . What are the eigenvalues and eigenvectors the of number operators? Are there any degeneracies? c) Confirm that the all commutation relations given on Ba-416 hold for the CA operators in the matrix representation. HINT: I gave up after doing {a0 , a†0 } = 1

and

{a0 , a†1 } = 0 .

156 Chapt. 24 Second Quantization 024 qfull 00300 2 5 0 moderate thinking: one-particle density matrix Extra keywords: See Ba-425 3. A quantity that turns out to be useful in studying non-interacting, spin 1/2 fermions in a box quantization system of volume V is the one-particle density matrix defined by Gs (~r − ~r ′ ) = hΦ0 |Ψs (~r )† Ψs (~r ′ )|Φ0 i , where |Φ0 i is the ground state and Ψs (~r )† =

X e−i~p·~r † √ ap~s V p ~

and

Ψs (~r ) =

X ei~p·~r √ ap~s V p ~

are the field creation and annihilation operators with − h set to 1 and where s is the z-quantum number for the spin state. Recall that the ground state has all the states occupied for |~ p | ≤ pf , where pf is the Fermi momentum. to a) Find an expression for Gs (0) using the approximation of a continuum of states. What, in fact, is this quantity? b) Show that Gs (~r − ~r ′ ) is given by Gs (~r − ~r ′ ) =

3n sin(x) − x cos(x) , 2 x3

where x ≡ pf |~r −~r ′ |, n is the mean density, and the approximation of a continuum of states has been used. HINT: The integrand is not isotropic in this case, but one can choose the z-axis for maximum simplicity. c) Now let us analyze the dimensionless Gs (~r − ~r ′ ) given by g(x) =

sin(x) − x cos(x) , x3

where only x greater than is meaningful, of course. First, what is the variation in x usually to be thought to be attributed to? Second, what are the small x and large x limiting forms of g(x): give the small x limiting form to 4th order in x. Third, give an approximate expression for the zeros of g(x). Fourth, sketch g(x). d) Now find a convergent iteration formula that allows you to solve for the zeros of g(x). Implement this formula in a computer code and compute the first ten zeros to good accuracy. 024 qfull 00400 2 5 0 moderate thinking: fermion pair correlation Extra keywords: See Ba-428–429 and Ar-527 4. The pair correlation function for spin 1/2 fermions in the ground state of a box quantization system is ( 1, for s 6= s′ ; ′ 9 gss′ (~r − ~r ) = 1 − 6 [sin(x) − x cos(x)]2 , for s = s′ , x where s is a spin coordinate label, x = pf |~r −~r ′ |, pf is the Fermi momentum, ~r is the point where a fermion has be removed, and ~r ′ is the point where one has measured the particle density just after the removal of the fermion (Ba-428–429). The expectation particle density density for a given is spin coordinate is (n/2)gss′ (~r − ~r ′ ), where n = N/V is the original expectation density of a spin state: N is the number of particles in the system and V is the system volume.

Chapt. 24 Second Quantization 157 a) Prove that (n/2)gss′ (~r − ~r ′ ) is properly scaled for s 6= s′ : i.e., that the integral of (n/2)gss′ (~r − ~r ′ ) over all volume is the right number of particles.

b) Now prove that (n/2)gss′ (~r − ~r ′ ) is properly scaled for s = s′ . HINT: You might have to look up the properties of the spherical Bessel functions—unless you know those like the back of your hand. 024 qfull 00450 2 5 0 moderate thinking: 2nd quantization potential Extra keywords: See Ba-434 5. The 2nd quantization operator for a two-body potential V (~r − ~r ′ ) between identical particles is V2nd

1X = 2 ′ ss

Z

d~r ′ d~r V (~r − ~r ′ )Ψs (~r )† Ψs′ (~r ′ )† Ψs′ (~r ′ )Ψs (~r ) ,

where the 1/2 prevents double counting in the integration, the sum is over all spin states, and the Ψ† ’s and Ψ’s are the field creation and annihilation operators. Prove that V2nd is correct by showing that a matrix element of V2nd with general, properly symmetrized, n particle states |Φ′ i and |Φi (i.e., hΦ′ |V2nd |Φi) is the same as the matrix element of the same states with the 1st quantization operator for n identical particles: i.e., V1st =

1 X V (~ri − ~rj ) , 2 ij,i6=j

where the 1/2 prevents double counting. HINTS: You should recall the effect of a field creation operator on a localized state of n − 1 particles: i.e., Ψs (~r )† |~r1 s1 , . . . , ~rn−1 sn−1 i =

√ n|~r1 s1 , . . . , rn−1 sn−1 , rn sn i

(Ba-419) Also recall localized state unit operator for a properly symmetrized states of n particles is X Z 1n = d~r1 . . . ~rn |~r1 s1 , . . . , ~rn sn ih~r1 s1 , . . . , ~rn sn | s1 ...sn

(Ba-421). 024 qfull 00500 3 5 0 tough thinking: fermion Coulomb exchange Extra keywords: See Ba-436 6. The exchange energy per electron for a ground state electron gas is given by E 9πne2 =− 2 N pf

Z

0

∞

[sin(x) − x cos(x)]2 dx , x5

where N is electron number, n is electron density, pf is the Fermi momentum, and x = pf r is a dimensionless radius for an integration over all space (Ba-436). a) First, let us analyze the integrand of the integral. Where are its zeros approximately for x ≥ 0? What is its small x behavior to 5th order in small x? What is its large x behavior? Sketch the integrand. b) Now evaluate the integral approximately or exactly by analytic means. HINT: An exact analytic integration must be possible, but probably one must use some special method. For approximate analytic integration, just do the best you can. c) Now evaluate the integral numerically to high accuracy. HINT: Simpson’s rule with double precision fortran works pretty well.

158 Chapt. 24 Second Quantization 024 qfull 00600 2 5 0 moderate thinking: two fermions in a box Extra keywords: (Ba-439:4) 7. Consider a box quantization system with volume V : the single-particle eigenstates recall are given by ei~p·~r , φs (~r ) = V where periodic boundary conditions have been imposed, − h has been set to 1, and s gives the spin coordinate. We will consider two spin 1/2 fermions in the box in state |1p~1 s1 , 1p~2 s2 i = ap†~2 s2 ap†~1 s1 |0i , where the a† ’s are creation operators and |0i is the vacuum state (Ba-417). anticommutation relations for the fermion creation and annihilation operators: {ap~s , ap†~′ s′ } = δp~p~′ δss′ ,

{ap~s , ap~′ s′ } = 0 ,

and

Recall the

{ap†~s , ap†~′ s′ } = 0 ,

where the a† ’s are again creation operators and the a’s are annihilation operators (Ba-417). a) Show that h1p~1 s1 , 1p~2 s2 |1p~1 s1 , 1p~2 s2 i = 1 − δp~1 p~2 δs1 s2 . What is the interpretation of this result? b) You are now given the operator ap†~s aq†~s′ aq~ ′ s′ ap~ ′ s . Write out the explicit expectation value of the operator for the |1p~1 s1 , 1p~2 s2 i state. HINTS: Remember to make use of the annihilation property a|0i = 0. There are a lot of tedious Kronecker deltas. c) One form of the 2nd quantization two-body interaction operator is v2nd

Z ′ ′ ′ 1 1 X X = d~r d~r ′ v(~r − ~r ′ )e−i(~p−~p )·~r e−i(~q−~q )·~r ap†~s aq†~s′ aq~ ′ s′ ap~ ′ s , 2V2 ′ ′ ′ pp qq ss

where V again is volume, v(~r − ~r ′ ) is the two-body potential function, and ap†~s aq†~s′ aq~ ′ s′ ap~ ′ s is the operator from the part (b) question. Making use of the part (b) answer express expectation value h1p~1 s1 , 1p~2 s2 |v2nd |1p~1 s1 , 1p~2 s2 i as simply as possible. Don’t assume that surface effects can be neglected. What is the use or significance of this matrix element? HINTS: The summations allow one to kill most of the Kroneckar deltas and get a pretty simple expression, but the explicit integrations are still there. d) Say the two fermions are in different spin states and the two-body potential v(~r − ~r ′ ) = C, where C is a constant. What is the expectation value from the part (c) answer in this case?

Chapt. 25 Klein-Gordon Equation

Multiple-Choice Problems 025 qmult 00100 1 4 4 easy deducto-memory: Klein-Gordan eqn Extra keywords: See the biography of Schrodinger and BJ-343 1. “Let’s play Jeopardy! For $100, the answer is: A relativistic quantum mechanical particle equation originally invented by Schr¨odinger (but never published by him) whose valid interpretation was provided by Pauli and Weisskopf.” a) b) c) d) e)

What What What What What

is is is is is

the the the the the

2nd Schr¨odinger equation, Alex? relativistic Sch¨ odinger equation, Alex? Pauli-Weisskopf equation, Alex? Klein-Gordon equation, Alex? small Scotsman equation, Alex?

025 qmult 00200 2 1 2 moderate memory: KG eqn development Extra keywords: See Ba-501 2. To develop the free-particle Klein-Gordon equation one assumes that the operators Eop = i− h ∂/∂t and pop = (− h /i)∇ apply in relativistic quantum mechanics and then applies the correspondence principle to the non-quantum-mechanical special relativity result: p a) E = (pc)2 + (mc2 )2 . 2 2 2 b) E 2 = (pc) p + (mc ) . 2 c) γ = 1/ 1 − β . d) E = mc2 . p e) ℓ = ℓproper 1 − β 2 .

025 qmult 00300 1 1 2 easy memory: antiparticles Extra keywords: See Ba-506 3. The free-particle Klein-Gordon equation leads to two energy eigen-solutions for each momentum eigenvalue. The two energies are equal in absolute value and opposite in sign in one interpretation. The negative energy solution can, in fact, be interpreted as a positive energy solution for: a) b) c) d) e)

a helium atom. an antiparticle. a wavicle. an anti-wavicle. an anti-wastrel.

025 qmult 00400 1 1 4 easy memory: KG density and current density Extra keywords: See Ba-503 and BJ-343 4. The density and current density that one obtains from the Klein-Gordon wave function are interpreted, respectively, as: a) anti-density and anti-current density. 159

160 Chapt. 25 Klein-Gordon Equation b) c) d) e)

improbability density and current density. current density and density in a relativistic reversal. expectation value charge density and charge current density. probability density and probability current density.

Full-Answer Problems 025 qfull 00300 2 5 0 moderate thinking: Lorentz transf. of KG eqn Extra keywords: See Ba-502 1. One form of the free-particle Klein-Gordon (KG) equation is "

1 ∂2 Kop Ψ(~r, t) = 2 2 − ∇2 + c ∂t



mc − h

2 #

Ψ(~r, t) = 0 ,

where Kop is here just an abbreviation for the Klein-Gordon equation operator and m is rest mass (Ba-501). This equation is supposed to be relativistically correct. This means that it must be the correct physics to apply to the particle in any inertial frame. But how does the solution Ψ(~r, t) change on transformation from one inertial frame to another? To find out, consider the special case of a transformation between frames S and S ′ , where the S ′ is moving at velocity β (in units of c) along the mutual x-axis of the two frames. The origins of the two frames were coincident at τ = τ ′ = 0 (where τ = ct: i.e., time in units of distance). The special Lorentz transformations in this case are x′ = γ(x − βτ ) ,

τ ′ = γ(τ − βx) ,

where the Lorentz factor

y′ = y ,

and

z′ = z ,

1 . γ= p 1 − β2

Now transform Kop Ψ(~r, t) = 0 to the primed system (i.e., write it terms of the ~r ′ and t′ ′ variables). Does the transformed equation Kop Ψ[~r = f (~r ′ , t′ ), t = f (~r ′ , t′ )] = 0 have the same form as the untransformed equation: i.e., does it look like "

1 ∂2 − ∇′2 + c2 ∂t′2



mc − h

2 #

Ψ(~r ′ , t′ )′ = 0

when Ψ[~r = f (~r ′ , t′ ), t = f (~r ′ , t′ )] is identified as Ψ(~r ′ , t′ )′ ? What kind of object is Ψ(~r, t): i.e., scalar, vector, tensor? HINT: Recall the chain rule.

Multiple-Choice Problems

Full-Answer Problems 030 qfull 00100 2 5 0 moderate thinking: eph Bose-Einstein condensate Extra keywords: Greiner et al. 2002, 415, 39 with Stoof review on p. 25

Chapt. 25 Klein-Gordon Equation 161 1. Go to the library 2nd floor reading room and find the 20002jan03 issue of Nature (it may have been placed under the display shelf) and read the commentary article by Stoof on page 25 about a quantum phase transition from a Bose-Einstein condensate to a Mott insulator. What do you think Stoof really means when he says in the superfluid state “atoms still move freely from one valley to the next”? NOTE: The instructor disavows any ability to completely elucidate this commentary or the research article by Greiner et al. it comments on. 030 qfull 00200 2 3 0 moderate thinking: eph quantum gravity well Extra keywords: reference: Nesvizhevsky et al. 2002, Nature, 413, 297 2. Is the gravity subject to quantum mechanical laws or is it somehow totally decoupled? Everyone really assumes that gravity is subject to quantum mechanical laws, but the assumption is not well verified experimentally: in fact it may never have been verified at all until now—not that I would know. The lack of experimental verification is because gravity is so infernally weak compared to other forces in microscopic experiments that it is usually completely negligible. Recently Nesviszhevsky et al. (2002, Nature, 413, 297) have reported from an experiment that a gravity well (at least part of the constraining potential is gravitational) does have quantized energy states. This appears to be the first time that such an experimental result has been achieved. It’s a wonderful result. Of course, if they hadn’t found quantization, it would have been a shock and most people would have concluded that the experiment was wrong somehow. Experiment may be the ultimate judge of theory, but experiment can certainly tell fibs for awhile. Go read Nesviszhevsky et al. in the 2nd floor reading room of the library: the relevant issue may be under the shelf. If a neutron in the theoretically predicted gravity well made a transition from the 1st excited state to the ground state and emitted a photon, what would be the wavelength of the photon? Could such a photon be measured? What classically does such a transition correspond to? 030 qfull 00300 1 5 0 easy thinking: eph quantum computing Extra keywords: Reference Seife, C. 2001, Science, 293, 2026 3. Read the article on quantum computing by Seife (2001, Science, 293, 2026) and make an estimate of how long it will be before for there is a quantum computer that solves a computational problem not solvable by a classical computer: I’m excepting, of course, any problems concerning quantum computer operation itself. Give your reasoning. All answers are right—and wrong—or in a superposition of those two states. My answer is 1 year. HINT: You can probably find the issue in the library, but there’s one in the physics lounge near the Britney issue. Primers on quantum computing can be found by going to http://www.physics.unlv.edu/~jeffery/images/science and clicking down through quantum mechanics and quantum computing. 030 qfull 00400 2 5 0 moderate thinking: eph C-70 diffraction Extra keywords: Reference Nairz et al. 2001, quant-ph/0105061 4. On the web go to the Los Alamos eprint archive: http://xxx.lanl.gov/ . There click on search and then on search for articles by Zeilinger under the quant-ph topic. Locate Nairz, Arndt, & Zeilinger 2001, quant-ph/0105061 and download it. This is article reports the particle diffraction for C70 (a fullerene). There should be great pictures on the web of fullerenes, but the best I could find were at http://www.sussex.ac.uk/Users/kroto/fullgallery.html and http://cnst.rice.edu/pics.html

162 Chapt. 25 Klein-Gordon Equation and these don’t have descriptions. Fullerenes are the largest particles ever shown to diffract: their size scale must be order-of a nanometer: 10 times ordinary atomic size. The article calls itself a verification of the Heisenberg uncertainty principle. In a general sense this is absolutely true since they verify the wave nature of particle propagation. But it isn’t a direct test of the formal uncertainty relation − h , σx σpx ≥ 2 where σx and σpx are standard deviations of x-direction position and momentum, respectively, for the wave function (e.g., Gr-18, Gr-108–110). Explain why it isn’t a direct test. HINTS: You should all have studied physical optics at some point. Essentially what formula are they testing? 030 qfull 00500 1 5 0 easy thinking: bulk and branes Extra keywords: Reference Arkani-Hamed, N., et al. 2002, Physics Today, February, 35 5. Go the library basement and read the article by Arkani-Hamed et al. (Physics Today, February, p. 35). Perhaps a 2nd reading would help or a course in particle physics. Anyway what is the bulk (not Hulk, bulk) and the brane (not Brain, brane)? 030 qfull 00600 1 5 0 easy thinking: sympathetic cooling Extra keywords: O’Hara & Thomas, 2001, Science, March 30, 291, 2556 6. Go to the library or the physics lounge and read O’Hara & Thomas ( 2001, Science, March 30, 291, p. 2556) on degenerate gases of bosons and fermions. What is sympathetic cooling?

Appendix 1 Mathematical Problems

Multiple-Choice Problems

Full-Answer Problems 031 qfull 00100 2 3 0 moderate math: Gauss summations 1. Gauss at the age of two proved various useful summation formulae. Now we can do this too maybe. a) Prove S0 (n) =

n X

1=n.

ℓ=1

HINT: This is really very easy. b) Prove S1 (n) =

n X

ℓ=

ℓ=1

n(n + 1) . 2

HINT: The trick is to add to every term in the sum its “complement” and then sum those 2-sums and divide by 2 to account for double counting. c) Prove S2 (n) =

n X

ℓ2 =

ℓ=1

n(n + 1)(2n + 1) . 6

HINT: A proof by induction works, but for that proof you need to know the result first and that’s the weak way. The stronger way is to reduce the problem to an already solved problem. Consider the general summation formula Sk (n) =

n X

ℓk .

ℓ=1

For each ℓ, you can construct a column of ℓk−1 ’s that is ℓ in height. Can you add up the values in the table that is made up of these columns in some way to get Sk (n). d) Prove S3 (n) =

n X ℓ=1

ℓ3 =

n2 (n + 1)2 . 4

HINT: This formula can be proven using the “complement” trick and the formulae of parts (b) and (c). It can also more tediously be solved by the procedure hinted at in part (c). Or, of course, induction will work.

163

164 Appendix 1 Mathematical Problems 031 qfull 00200 1 3 0 easy math: uniqueness of power series 2. Power series are unique. a) Prove that coefficients ak of the power series P (x) =

∞ X

ak xk

k=0

are unique choices given that the series is convergent of course. HINT: The mth derivative of P (x) evaluated at x = 0 can have only one value. b) Prove that coefficients akℓ of the double power series P (x, y) =

∞,∞ X

akℓ xk y ℓ

k=0,ℓ=0

are unique choices given that the series is convergent of course. HINT: Mutatis mutandis. 031 qfull 00300 2 5 0 moderate thinking: Leibniz’s formula (Ar-558) proof 3. Prove Leibniz’s formula (Ar-558) n   dn (f g) X n dk f dn−k g = dxn k dxk dxn−k k=0

by induction. 031 qfull 00400 2 5 0 moderate thinking: integrals of type xe**-2 4. In evaluating anything that depends on a Gaussian distribution (e.g., the Maxwell-Boltzmann distribution of classical statistical mechanics), one frequently has to evaluate integrals of the type Z ∞

In =

2

xn e−λx dx ,

0

where n is an odd positive integer. a) Solve for I1 . b) Obtaining the general formula for In is now trivial with a magic trick. Act on I1 with the operator  (n−1)/2 d − . dλ c) From the general formula evaluate I1 I3 , I5 , and I7 . 031 qfull 01000 2 5 0 moderate thinking: 5. In understanding determinants some permutation results must be proven. The proofs are expected to be cogent and memorable rather than mathematical rigorous. a) Given n objects, prove that there are n! permutations for ordering them in a line. b) If you interchange any two particles in a given permutation, you get another permutation. Let’s call that action an exchange. If you exchange nearest nearest neighbors, let’s call that a nearest neighbor or NN exchange. Prove that any exchange requires an odd number of NN exchanges.

Appendix 1 Mathematical Problems 165 c) Permutations have definite parity. This means that going from one definite permutation to another definite permutation by any possible series of NN exchanges (i.e., by any possible path) will always involve either an even number of NN exchanges or an odd number: i.e., if one path is even/odd, then any other path is even/odd. Given that definite parity is true prove that any path of NN exchanges from a permutation that brings you back to that permutation (i.e., a closed path) has an even number of NN exchanges. d) Now we have to prove definite parity exists. Say there is a fiducial permutation which by definition we say has even parity. If definite parity exists, then every other permutation is definitely even or odd relative to the fiducial permutation. If n = 1, does definite parity hold in this trivial case? For n ≥ 2, prove that definite parity holds. HINTS: It suffices to prove that going from the fiducial permutation to any other permutation always involves a definite even or odd path since the fiducial permutation is arbitrary. Proof by induction might be the best route. I can’t see how brief word arguments can be avoided. e) Now prove for n ≥ 2 that there are an equal number of even and odd permutations. HINT: Consider starting with an even permutation and systematically by an NN exchange path going through all possible permutations. Then start with an odd permutation and follow the same NN exchange path.

Appendix 2 Quantum Mechanics Equation Sheet Note: This equation sheet is intended for students writing tests or reviewing material. Therefore it neither intended to be complete nor completely explicit. There are fewer symbols than variables, and so some symbols must be used for different things. 1 Constants not to High Accuracy Constant Name

Symbol

Derived from CODATA 1998

Bohr radius

aBohr =

Boltzmann’s constant

k

Compton wavelength

λCompton =

Electron rest energy Elementary charge squared

m e c2 e2

λCompton 2πα h me c

e2 α= − hc 2 − h 2me 2 − h me h − h hc − hc 1 ERyd = me c2 α2 2

Fine Structure constant Kinetic energy coefficient

Planck’s constant Planck’s h-bar

Rydberg Energy

= 0.529 ˚ A = 0.8617 × 10−6 eV K−1 = 1.381 × 10−16 erg K−1 = 0.0246 ˚ A = 5.11 × 105 eV = 14.40 eV ˚ A = 1/137.036 = 3.81 eV ˚ A2 = 7.62 eV ˚ A2 = 4.15 × 10−15 eV = 6.58 × 10−16 eV = 12398.42 eV ˚ A = 1973.27 eV ˚ A = 13.606 eV

2 Some Useful Formulae Leibniz’s formula

n   dn (f g) X n dk f dn−k g = dxn k dxk dxn−k k=0

Normalized Gaussian

  1 (x − hxi)2 P = √ exp − 2σ 2 σ 2π

3 Schr¨ odinger’s Equation  p2 ∂Ψ(x, t) HΨ(x, t) = + V (x) Ψ(x, t) = i− h 2m ∂t 

 p2 Hψ(x) = + V (x) ψ(x) = Eψ(x) 2m 

 ∂Ψ(~r , t) p2 + V (~r ) Ψ(~r , t) = i− h HΨ(~r , t) = 2m ∂t 

166

∂ H|Ψi = i− h |Ψi ∂t

Appendix 2 Quantum Mechanics Equation Sheet 167  p2 Hψ(~r ) = + V (~r ) ψ(~r ) = Eψ(~r ) 2m 

H|ψi = E|ψi

4 Some Operators p=

H=

− h ∂ i ∂x

2 − p2 h ∂2 + V (x) + V (x) = − 2m 2m ∂x2

p=

H=

∇=x ˆ

2 2 ∂ p2 = −− h ∂x2

− h ∇ i

2 p2 = −− h ∇2

2 − p2 h + V (~r ) = − ∇2 + V (~r ) 2m 2m

∂ ∂ ∂ 1 ∂ 1 ∂ ∂ + yˆ + zˆ = rˆ + θˆ + θˆ ∂x ∂y ∂z ∂r r ∂θ r sin θ ∂θ

∂2 ∂2 ∂2 1 ∂ ∇ = + + = 2 2 2 2 ∂x ∂y ∂z r ∂r 2

    1 ∂ 1 ∂ ∂2 2 ∂ r + 2 sin θ + 2 2 ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2

5 Kronecker Delta and Levi-Civita Symbol

δij =



1, i = j; 0, otherwise

εijk =

εijk εiℓm = δjℓ δkm − δjm δkℓ

1, ijk cyclic; −1, ijk anticyclic; 0, if two indices the same.

(

(Einstein summation on i)

6 Time Evolution Formulae General

dhAi = dt

Ehrenfest’s Theorem

|Ψ(t)i =



∂A ∂t



1 + − hi[H(t), A]i h

dh~r i 1 = h~ pi dt m X j

and

dh~ pi = −h∇V (~r )i dt

− cj (0)e−iEj t/ h |φj i

7 Simple Harmonic Oscillator (SHO) Formulae 1 V (x) = mω 2 x2 2

2 − h ∂2 1 − + mω 2 x2 2m ∂x2 2

!

ψ = Eψ

168 Appendix 2 Quantum Mechanics Equation Sheet

β=

r

mω − h

ψn (x) =

2 2 β 1/2 1 √ Hn (βx)e−β x /2 1/4 n π 2 n!

H0 (βx) = H0 (ξ) = 1

En =

  1 − hω n+ 2

H1 (βx) = H1 (ξ) = 2ξ

2

H3 (βx) = H3 (ξ) = 8ξ 3 − 12ξ

H2 (βx) = H2 (ξ) = 4ξ − 2

8 Position, Momentum, and Wavenumber Representations p=− hk

Ekinetic = ET =

|Ψ(p, t)|2 dp = |Ψ(k, t)|2 dk xop = x

xop

pop =

− h ∂ =− i ∂p

− h ∂ i ∂x

δ(x) =

∞

−∞

 −  h ∂ Q x, ,t i ∂x

Z

∞ −∞

−

∞

Ψ(x, t) =

Ψ(k, t) =

−

all space

∞

Z

∞

Ψ(k, t)

eikx dk (2π)1/2

Ψ(x, t)

e−ikx dx (2π)1/2

e−i~p·~r/ h 3 d r (2π− h )3/2

~

eik·~r d3 k Ψ(~k , t) (2π)3/2 all space

Ψ(~r , t) =

Z

Ψ(~k , t) =

Z

−

Ψ(~r , t)

Z

−∞

ei~p·~r/ h 3 Ψ(~ p , t) Ψ(~r , t) = d p (2π− h )3/2 all space Z

eikx dk 2π

−∞

−

∞

Z

Ψ(~ p , t) =

momentum representation

δ(x) =

e−ipx/ h dx Ψ(x, t) Ψ(p, t) = (2π− h )1/2 −∞ Z

position representation

−

eipx/ h dp 2π− h

eipx/ h Ψ(x, t) = Ψ(p, t) dp (2π− h )1/2 −∞ Z

Ψ(k, t) √ − h

Ψ(p, t) =

 −  h ∂ Q − , p, t i ∂p

pop = p

Z

2 − h k2 2m

~

Ψ(~r , t) all space

e−ik·~r 3 d r (2π)3/2

9 Commutator Formulae   X X X  ai bj [Ai , bj ] bj Bj  = ai Ai ,

[A, BC] = [A, B]C + B[A, C]

i

if

[B, [A, B]] = 0

[x, p] = i− h

then

[x, f (p)] = i− h f ′ (p)

j

i,j

[A, F (B)] = [A, B]F ′ (B) [p, g(x)] = −i− h g ′ (x)

Appendix 2 Quantum Mechanics Equation Sheet 169 [a, a† ] = 1

[N, a† ] = a†

[N, a] = −a

10 Uncertainty Relations and Inequalities σx σp = ∆x∆p ≥

− h 2

σQ σQ = ∆Q∆R ≥

σH ∆tscale time = ∆E∆tscale time ≥

1 |hi[Q, R]i| 2

− h 2

11 Probability Amplitudes and Probabilities P (dx) = |Ψ(x, t)|2 dx

Ψ(x, t) = hx|Ψ(t)i

P (i) = |ci (t)|2

ci (t) = hφi |Ψ(t)i

12 Spherical Harmonics 1 Y0,0 = √ 4π

Y1,0 =

2 L2 Yℓm = ℓ(ℓ + 1)− h Yℓm

0 s



3 4π

1/2

cos(θ)

Y1,±1 = ∓

Lz Yℓm = m− h Yℓm

1 p

2 d

|m| ≤ ℓ

3 f



3 8π

1/2

sin(θ)e±iφ

m = −ℓ, −ℓ + 1, . . . , ℓ − 1, ℓ

4 g

5 h

6 i

13 Hydrogenic Atom ψnℓm = Rnℓ (r)Yℓm (θ, φ)

az =



a Z

me mreduced



a0 =

ℓ = 0, 1, 2, . . . , n − 1

− h λC = me cα 2πα

e2 α= − hc

  1 r 1 −3/2 1− e−r/(2aZ ) R20 = √ aZ 2 aZ 2

−3/2 −r/aZ

e

R10 = 2aZ

ℓ≤n−1

1 −3/2 r −r/(2aZ ) R21 = √ aZ e aZ 24

Rnℓ = − Lq (x) = ex



(

d dx

Ljq (x)

q =

2 naZ

3

e−x xq 

d dx

j

(n − ℓ − 1)! 2n[(n + ℓ)!]3




)1/2

e−ρ/2 ρℓ L2ℓ+1 n+ℓ (ρ)

ρ=

2r nrZ

Rodrigues’s formula for the Laguerre polynomials

Lq (x)

Associated Laguerre polynomials

... ...

170 Appendix 2 Quantum Mechanics Equation Sheet

hrinℓm =

 aZ  2 3n − ℓ(ℓ + 1) 2

Nodes = (n − 1) − ℓ

not counting zero or infinity

Z 2 mreduced Z 2 mreduced Z 2 mreduced 1 = −ERyd 2 = −13.606 2 eV En = − me c2 α2 2 2 n me n me n me 14 General Angular Momentum Formulae [Ji , Jj ] = i− h εijk Jk

2

J 2 |jmi = j(j + 1)− h |jmi

J{ x } = y

1 2 1 2i

)

[Jf i , Jgj ] = δf g i− h εijk Jk

J± = J1± + J2±

Jz |jmi = m− h |jmi

p J± |jmi = − h j(j + 1) − m(m ± 1)|jm ± 1i

J± = Jx ± iJy (

[J 2 , J~ ] = 0

(Einstein summation on k)

† J± J± = J∓ J± = J 2 − Jz (Jz ± − h)

(J+ ± J− )

J~ = J~1 + J~2

|j1 j2 jmi =

J 2 = J12 + J22 + J1+ J2− + J1− J2+ + 2J1z J2z

X

m1 m2 ,m=m1 +m2

jX 1 +j2

|j1 − j2 | ≤ j ≤ j1 + j2

|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 jmij1 j2 jmi

(2j + 1) = (2j1 + 1)(2j2 + 1)

|j1 −j2 |

15 Spin 1/2 Formulae  − h 0 Sx = 2 1

1 0



1 |±ix = √ (|+i ± |−i) 2

−i 0



  − h 1 0 Sz = 2 0 −1

1 |±iy = √ (|+i ± i|−i) 2

|±iz = |±i

1 | + −i = √ (|1, +i|2, −i ± |1, −i|2, +i) 2

| + +i = |1, +i|2, +i

σx =

 − h 0 Sy = 2 i



0 1 1 0



σy =



0 −i i 0



σz =



| − −i = |1, −i|2, −i

1 0 0 −1



Appendix 2 Quantum Mechanics Equation Sheet 171 σi σj = δij + iεijk σk

[σi , σj ] = 2iεijk σk

{σi , σj } = 2δij

~ · ~σ )(B ~ · ~σ ) = A ~·B ~ + i(A ~ × B) ~ · ~σ (A ~ ·n d(S ˆ) i ~ ~·n ·α ˆ, S ˆ] = − − [S dα h eixA = 1 cos(x) + iA sin(x)

~ ~ ~ ~ ·n S ˆ = e−iS·~α S ·n ˆ 0 eiS·~α

~

|ˆ n± i = e−iS·~α |ˆ z± i

e−i~σ ·~α/2 = 1 cos(x) − i~σ · α ˆ sin(x)

if A2 = 1

σi f (σj ) = f (σj )σi δij + f (−σj )σi (1 − δij ) µBohr =

e− h = 0.927400915(23) × 10−24 J/T = 5.7883817555(79) × 10−5 eV/T 2m   α g = 2 1+ + . . . = 2.0023193043622(15) 2π

~ L µorbital = −µBohr − ~ h

~ S µ ~ spin = −gµBohr − h ~ Hµ = −~µ · B

~µtotal = ~µorbital + ~µspin = −µBohr

Hµ = µBohr Bz

~ + g S) ~ (L − h

(Lz + gSz ) − h

16 Time-Independent Approximation Methods H = H (0) + λH (1)

H (1) |ψn(m−1) i(1 − δm,0 ) + H (0) |ψn(m) i =

X

|ψn1st i = |ψn(0) i + λ En1st

|ψi = N (λ)

m X ℓ=0

D

(0)

(0)

ψk |H (1) |ψn (0)

(0)

E

En2nd = En(0) + λ ψn(0) |H (1) |ψn(0) + λ2

E(φ) =

k=0

E (m−ℓ) |ψn(ℓ) i

En − Ek all k, k6=n D E = En(0) + λ ψn(0) |H (1) |ψn(0) D

∞ X

hφ|H|φi hφ|φi

E

λk |ψn(k) i

|ψn(ℓ>0) i =

∞ X

m=0, m6=n

(0)

|ψk i

X

all k, k6=n

D E (0) (1) (0) 2 ψk |H |ψn

δE(φ) = 0

(0)

(0)

En − Ek

anm |ψn(0) i

172 Appendix 2 Quantum Mechanics Equation Sheet Hkj = hφk |H|φj i

H~c = E~c

17 Time-Dependent Perturbation Theory π=

Z

∞

−∞

sin2 (x) dx x2

2π Γ0→n = − |hn|Hperturbation|0i|2 δ(En − E0 ) h 18 Interaction of Radiation and Matter ~ ~ op = − 1 ∂ Aop E c ∂t

~ op = ∇ × A ~ op B

19 Box Quantization kL = 2πn,

n = 0, ±1, ±2, . . .

k=

2πn L

dNstates = g

∆kcell =

2π L

3 ∆kcell =

(2π)3 V

k 2 dk dΩ (2π)3 /V

20 Identical Particles 1 |a, bi = √ (|1, a; 2, bi ± |1, b; 2, ai) 2 1 ψ(~r1 , ~r 2 ) = √ (ψa (~r 1 )ψb (~r 2 ) ± ψb (~r 1 )ψa (~r 2 )) 2 21 Second Quantization

[ai , a†j ] = δij

[ai , aj ] = 0

{ai , a†j } = δij

{ai , aj } = 0

Ψs (~r )† =

[Ψs (~r ), Ψs′ (~r ′ )]∓ = 0

(a† )N1 (a† )Nn . . . √1 |0i |N1 , . . . , Nn i = √n Nn ! N1 !

[a†i , a†j ] = 0

|N1 , . . . , Nn i = (a†n )Nn . . . (a†1 )N1 |0i

{a†i , a†j } = 0

X e−i~p·~r † √ ap~s V p ~

Ψs (~r ) =

[Ψs (~r )† , Ψs′ (~r ′ )† ]∓ = 0

X ei~p·~r √ ap~s V p ~

[Ψs (~r ), Ψs′ (~r ′ )† ]∓ = δ(~r − ~r ′ )δss′

Appendix 2 Quantum Mechanics Equation Sheet 173 1 |~r1 s1 , . . . , ~rn sn i = √ Ψsn (~r n )† . . . Ψsn (~r n )† |0i n! √ Ψs (~r )† |~r1 s1 , . . . , ~rn sn i n + 1|~r1 s1 , . . . , ~rn sn , ~rsi |Φi = 1n =

X Z

Z

d~r1 . . . d~rn Φ(~r1 , . . . , ~rn )|~r1 s1 , . . . , ~rn sn i

d~r1 . . . d~rn |~r1 s1 , . . . , ~rn sn ih~r1 s1 , . . . , ~rn sn |

s1 ...sn

N=

X

ap†~s ap~s

T =

p ~s

ρs (~r ) = Ψs (~r )† Ψs (~r )

N=

Gs (~r − ~r ′ ) = v2nd

ss

v2nd =

Z

XZ

d~r ρs (~r )

T =

1 X 2m s

Z

d~r ∇Ψs (~r )† · ∇Ψs (~r )

 1  Ψs (~r )† ∇Ψs (~r ) − Ψs (~r )∇Ψs (~r )† 2im gss′ (~r − ~r ′ ) = 1 − δss′

Gs (~r − ~r ′ )2 (n/2)2

d~rd~r ′ v(~r − ~r ′ )Ψs (~r )† Ψs′ (~r ′ )† Ψs′ (~r ′ )Ψs (~r )

1 X X vp~−~p ′ δp~+~q,~p′ +~q′ ap†~s aq†~s′ aq~ ′ s′ ap~ ′ s 2V ′ ′ ′

vp~−~p ′ =

pp qq ss

Z

d~r e−i(~p−~p

′

)·~ r

22 Klein-Gordon Equation p E = p 2 c2 + m 2 c4 " ρ= 1 c2

1n

n=1

p ~s

3n sin(x) − x cos(x) 2 x3

1X = 2 ′

∞ X

X p2 † a ap~s 2m p~s

s

~js (~r ) =

1 = |0ih0| +

1 c2

# " 2  2 − ∂ h i− h Ψ(~r, t) = ∇ + m2 c2 Ψ(~r, t) ∂t i

1 ∂2 − ∇2 + c2 ∂t2



  ∂Ψ∗ i− h ∗ ∂Ψ Ψ − Ψ 2mc2 ∂t ∂t

mc − h

2 #

Ψ(~r, t) = 0

− ~j = h (Ψ∗ ∇Ψ − Ψ∇Ψ∗ ) 2im

# "  2 2 − ∂ e h 2 2 ~ + m c Ψ(~r, t) i− h − eΦ Ψ(~r, t) = ∇− A ∂t i c − Ψ+ (~ p, E) = ei(~p·~r−Et)/ h

− Ψ− (~ p, E) = e−i(~p·~r−Et)/ h

v(~r )

Appendix 3 Multiple-Choice Problem Answer Tables Note: For those who find scantrons frequently inaccurate and prefer to have their own table and marking template, the following are provided. I got the template trick from Neil Huffacker at University of Oklahoma. One just punches out the right answer places on an answer table and overlays it on student answer tables and quickly identifies and marks the wrong answers

Answer Table for the Multiple-Choice Questions a

b

c

d

e

a

b

c

d

e

1.

O

O

O

O

O

6.

O

O

O

O

O

2.

O

O

O

O

O

7.

O

O

O

O

O

3.

O

O

O

O

O

8.

O

O

O

O

O

4.

O

O

O

O

O

9.

O

O

O

O

O

5.

O

O

O

O

O

10.

O

O

O

O

O

174

Appendix 3 Multiple-Choice Problem Answer Tables 175

Answer Table for the Multiple-Choice Questions a

b

c

d

e

a

b

c

d

e

1.

O

O

O

O

O

11.

O

O

O

O

O

2.

O

O

O

O

O

12.

O

O

O

O

O

3.

O

O

O

O

O

13.

O

O

O

O

O

4.

O

O

O

O

O

14.

O

O

O

O

O

5.

O

O

O

O

O

15.

O

O

O

O

O

6.

O

O

O

O

O

16.

O

O

O

O

O

7.

O

O

O

O

O

17.

O

O

O

O

O

8.

O

O

O

O

O

18.

O

O

O

O

O

9.

O

O

O

O

O

19.

O

O

O

O

O

10.

O

O

O

O

O

20.

O

O

O

O

O

176 Appendix 3 Multiple-Choice Problem Answer Tables

Answer Table for the Multiple-Choice Questions a

b

c

d

e

a

b

c

d

e

1.

O

O

O

O

O

16.

O

O

O

O

O

2.

O

O

O

O

O

17.

O

O

O

O

O

3.

O

O

O

O

O

18.

O

O

O

O

O

4.

O

O

O

O

O

19.

O

O

O

O

O

5.

O

O

O

O

O

20.

O

O

O

O

O

6.

O

O

O

O

O

21.

O

O

O

O

O

7.

O

O

O

O

O

22.

O

O

O

O

O

8.

O

O

O

O

O

23.

O

O

O

O

O

9.

O

O

O

O

O

24.

O

O

O

O

O

10.

O

O

O

O

O

25.

O

O

O

O

O

11.

O

O

O

O

O

26.

O

O

O

O

O

12.

O

O

O

O

O

27.

O

O

O

O

O

13.

O

O

O

O

O

28.

O

O

O

O

O

14.

O

O

O

O

O

29.

O

O

O

O

O

15.

O

O

O

O

O

30.

O

O

O

O

O

Appendix 3 Multiple-Choice Problem Answer Tables 177

NAME: Answer Table for the Multiple-Choice Questions a

b

c

d

e

a

b

c

d

e

1.

O

O

O

O

O

21.

O

O

O

O

O

2.

O

O

O

O

O

22.

O

O

O

O

O

3.

O

O

O

O

O

23.

O

O

O

O

O

4.

O

O

O

O

O

24.

O

O

O

O

O

5.

O

O

O

O

O

25.

O

O

O

O

O

6.

O

O

O

O

O

26.

O

O

O

O

O

7.

O

O

O

O

O

27.

O

O

O

O

O

8.

O

O

O

O

O

28.

O

O

O

O

O

9.

O

O

O

O

O

29.

O

O

O

O

O

10.

O

O

O

O

O

30.

O

O

O

O

O

11.

O

O

O

O

O

31.

O

O

O

O

O

12.

O

O

O

O

O

32.

O

O

O

O

O

13.

O

O

O

O

O

33.

O

O

O

O

O

14.

O

O

O

O

O

34.

O

O

O

O

O

15.

O

O

O

O

O

35.

O

O

O

O

O

16.

O

O

O

O

O

36.

O

O

O

O

O

17.

O

O

O

O

O

37.

O

O

O

O

O

18.

O

O

O

O

O

38.

O

O

O

O

O

19.

O

O

O

O

O

39.

O

O

O

O

O

20.

O

O

O

O

O

40.

O

O

O

O

O

178 Appendix 3 Multiple-Choice Problem Answer Tables

Answer Table

Name:

a

b

c

d

e

a

b

c

d

e

1.

O

O

O

O

O

31.

O

O

O

O

O

2.

O

O

O

O

O

32.

O

O

O

O

O

3.

O

O

O

O

O

33.

O

O

O

O

O

4.

O

O

O

O

O

34.

O

O

O

O

O

5.

O

O

O

O

O

35.

O

O

O

O

O

6.

O

O

O

O

O

36.

O

O

O

O

O

7.

O

O

O

O

O

37.

O

O

O

O

O

8.

O

O

O

O

O

38.

O

O

O

O

O

9.

O

O

O

O

O

39.

O

O

O

O

O

10.

O

O

O

O

O

40.

O

O

O

O

O

11.

O

O

O

O

O

41.

O

O

O

O

O

12.

O

O

O

O

O

42.

O

O

O

O

O

13.

O

O

O

O

O

43.

O

O

O

O

O

14.

O

O

O

O

O

44.

O

O

O

O

O

15.

O

O

O

O

O

45.

O

O

O

O

O

16.

O

O

O

O

O

46.

O

O

O

O

O

17.

O

O

O

O

O

47.

O

O

O

O

O

18.

O

O

O

O

O

48.

O

O

O

O

O

19.

O

O

O

O

O

49.

O

O

O

O

O

20.

O

O

O

O

O

50.

O

O

O

O

O

21.

O

O

O

O

O

51.

O

O

O

O

O

22.

O

O

O

O

O

52.

O

O

O

O

O

23.

O

O

O

O

O

53.

O

O

O

O

O

24.

O

O

O

O

O

54.

O

O

O

O

O

25.

O

O

O

O

O

55.

O

O

O

O

O

26.

O

O

O

O

O

56.

O

O

O

O

O

27.

O

O

O

O

O

57.

O

O

O

O

O

28.

O

O

O

O

O

58.

O

O

O

O

O

29.

O

O

O

O

O

59.

O

O

O

O

O

30.

O

O

O

O

O

60.

O

O

O

O

O